ENGINEERING  LIBRARY 


ESSENTIALS 

IN  THE 
THEORY  OF  FRAMED  STRUCTURES 


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ESSENTIALS 

IN  THE 
THEORY  OF  FRAMED  STRUCTURES 


BY 


w 


CHARLES  A.  ELLIS,  A.  B. 

VICE-PRESIDENT THE  STRAUSS  BASCULE  BRIDGE  CO.     WITH  AMERICAN  BRIDGE  COMPANY 

IQOI-IQOS.        ASST.   PROFESSOR  OF  CIVIL  ENGINEERING,  UNIVERSITY  OF  MICHIGAN 

1008-1912.        DESIGNING  ENGINEER,  DOMINION  BRIDGE  COMPANY   1912-1914 

PROFESSOR  OF  STRUCTURAL  ENGINEERING,  UNIVERSITY  OF   ILLINOIS 

I9I4-I92I 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 

NEW  YORK:  370  SEVENTH  AVENUE 

LONDON:  6  &  8  BOUVERIE  ST.,  E.  C.  4 

1922 


OEM. 

uw 


ENGINEERING  LIBRARY 

COPYRIGHT,  IQ22,  BY  THE 
MCGRAW-HILL  BOOK  COMPANY,  INC. 


PRESS    YORK   PA 


PREFACE 

This  book,  designed  primarily  for  class-room  use,  contains  no 
new  principles.  The  same  fundamental  ideas  to  be  found  in 
many  other  books  are  here  set  forth  by  a  treatment  so  different 
from  the  orthodox  method  that  a  word  of  explanation  seems 
justifiable. 

There  is  a  growing  tendency  in  engineering  schools  to 
instruct  rather  than  to  educate;  to  pour  in  rather  than  to  draw 
out ;  to  feed  the  mind  with  memoranda  until  it  becomes  sluggish 
for  lack  of  exercise  in  the  intellectual  realms  of  the  imagination 
and  reason.  This  tendency  is,  to  a  large  extent,  the  result  of 
current  text-book  and  class-room  methods  which  present 
mathematical  principles  in  generalities  first  and  particulars  last. 
This  order  is  undesirable  for  two  reasons:  First,  it  is  not  the 
sequence  most  natural  for  the  grasp  of  the  human  intellect, 
for  the  history  of  any  science  will  show  that  its  development 
has  been  accomplished  by  progress  from  the  particular  to  the 
general.  Second,  if  generalities  are  considered  first,  the  princi- 
ple is  unfolded  to  the  class  by  deriving  a  formula.  Once 
developed,  this  formula  is  memorized,  and  all  subsequent 
problems  are  solved  by  merely  making  numerical  substitutions 
for  the  various  symbols.  Thus,  the  student  gets  but  a  passing 
glance  at  the  fundamental  idea  and  a  meager  conception  of  the 
principle  involved.  This  statement  can  be  proved  very  easily 
by  asking  any  group  of  students  who  can  glibly  recite : 

f-My  Me 

f~    ~T  I 

to  determine  the  fiber  stress  at  a  certain  point  in  a  given  beam 
without  using  a  formula.  This  experiment  was  tried  in  nearly 
one  thousand  cases  with  the  result  that  about  93  per  cent  of 
the  students  were  found  to  be  helpless  without  the  formula. 


M171603 


VI  PREFACE 

This  instance  is  more  fully  discussed  by  the  author  in 
an  article  "  Reinforced  Concrete  Theory  Without  the  Use  of 
Formulas,"  Bulletin  of  the  Society  for  the  Promotion  of 
Engineering  Education,  Vol.  VIII,  p.  380. 

In  this  book  each  principle  is  introduced  by  the  use  of  illus- 
trative numerical  problems,  some  of  which  will  lead  the  student 
into  a  pitfall  from  which  he  should  be  forced  to  extricate  himself 
without  the  instructor's  help.  Experience  is  a  splendid  teacher. 
There  is  no  opportunity  for  the  student  to  dodge  the  funda- 
mental idea  or  the  physical  conception  of  the  problem  by 
resorting  to  a  formula,  for  none  is  given  except  at  the  end  of  the 
discussion  when  the  general  case  is  considered.  Even  then 
the  general  expression  is  frequently  left  to  be  developed  by  the 
student.  The  experienced  teacher  who  has  had  considerable 
practical  experience  will,  no  doubt,  feel  more  at  home  with  this 
method  of  treatment  than  will  the  younger  instructor  who  is 
closely  confined  to  his  text  and  who  feels  that  a  knowledge  of 
the  course  is  of  more  importance  to  the  student  than  the 
development  of  mental  power. 

The  major  portion  of  the  book  may  be  read  intelligently  by 
any  student  who  has  a  working  knowledge  of  arithmetic,  algebra 
and  geometry.  Very  little  reference  (except  in  Chapter  V) 
is  made  to  the  calculus.  Practically  every  other  textbook 
makes  use  of  formal  calculus  in  the  treatment  of  maximum 
stresses  for  moving  loads.  Of  course,  this  method  is  entirely 
unnecessary.  It  is  true  that  the  fundamental  concept  of  the 
differential  calculus  is  ever  present  in  the  treatment  of  maximum 
stresses  in  connection  with  influence  lines  as  given  in  Chapter 
IV;  but  the  average  college  student  who  can  differentiate  and 
integrate  will  probably  not  recognize  the  calculus,  so  why  risk 
confusing  him  by  any  reference  whatever  to 

dy 
dx 

Some  new  material  will  be  found  in  Chapter  III  relative  to 
wind  reactions  of  a  mill-binding  bent.  A  treatment  of  deflec- 
tions by  the  area-moment  method,  more  comprehensive  than  is 
usually  found  in  American  textbooks,  appears  in  Chapter  V. 


PREFACE  Vll 

A  new  short  method  for  computing  the  stresses  in  a  swing  span 
is  given  in  Chapter  VIII. 

The  author  recommends  that  the  teaching  schedule  be 
arranged  to  include  one  three-hour  period  a  week  in  addition 
to  the  two  or  three  regular  recitation  periods.  These  three- 
hour  periods  should  be  used  for  written  tests  in  which  the 
student  is  given  an  opportunity  to  solve  problems  which  are 
just  ahead  of  the  textbook  assignments,  or  which  were  con- 
sidered so  far  back  in  the  student's  course  that  he  has  forgotten 
the  solution;  the  idea  being  to  give  the  student  some  opportun- 
ity for  independent  thought.  For  example:  Suppose  that 
during  a  certain  week  the  textbook  assignments  are  in  Sec. 
Ill,  Chapter  IV;  and  that  the  stress  in  a  diagonal  of  a  parallel 
chord  truss  has  been  discussed  in  the  recitation  period — the 
student  having  solved  several  problems  of  this  character.  The 
following  questions  are  recommended  for  the  three-hour  period 
of  that  week : 

1.  Prove  that  the  sum  of  the  three  angles  of  the  triangle 
equals  two  right  angles. 

2.  Compute    the    maximum    tensile   stress  in  the  number 
"l^La  of  the  curved  chord  truss  (Fig.  117)  for  an  E-6o  train  load. 

Any  simple  fundamental  problem  in  algebra,  calculus  or 
mechanics  will  serve  the  purpose  equally  as  well  as  the  problem 
in  geometry.  If  the  student  has  mastered  the  principle  of 
maximum  stress  in  the  diagonal  of  a  parallel  chord  truss  he 
should  be  given  the  opportunity  of  doing  some  independent 
thinking  on  the  principle  next  in  order,  instead  of  being  asked 
to  memorize  that  principle  from  the  text  in  preparation  for 
recitation. 

The  instructor  who  expects  that  the  student  will  submit  100 
per  cent  papers  on  an  examination  of  this  sort  will  be  sadly 
disappointed;  and  the  student  who  is  not  accustomed  to  this 
method  of  conducting  tests  will  at  first  feel  that  he  is  being 
treated  rather  roughly;  but  the  mental  development  which 
results  from  tests  of  this  character  is  of  more  value  to  the 
student  than  high  grades. 

The  major  portion  of  this  book  has  been  used  for  several 
years  as  preprints  in  the  author's  classes  and  he  is  indebted 


PREFACE 

to  former  students  for  many  helpful  suggestions.  The  author 
wishes  also  to  acknowledge  his  indebtedness  to  Professor  Geo. 
E.  Beggs  for  permission  to  use  his  table  of  equivalent  uniform 
loads. 

CHARLES  A.  ELLIS. 
CHICAGO,  ILLINOIS, 
Sept.  29,  192 1. 


CONTENTS 

PAGE 


PREFACE  ..............................  v 

CHAPTER  I 

SECTION  I.    INTRODUCTION 

1.  Mechanics  ...........................  x 

2.  Methods  and  Mental  Attitude  ..................  2 

3.  Forces  ............................  4 

4.  The  Three  Elements  of  a  Force   ................    .  5 

5.  Definitions  ..........................  ,5 

SECTION  II.    RESOLUTION  AND  COMBINATION  OF  FORCES 

6.  Parallelogram  of  Forces  .....................  6 

7.  Triangle  of  Forces  .......................  g 

8.  Magnitude-direction  Diagram  ........    ..........  g 

g.  Rectangular  Components  ..........    ..........  10 

10.  Moment  of  a  Force  .......................  n 

SECTION   III.    THE   THREE   FUNDAMENTAL   STATEMENTS   OR   EQUATIONS  OF 

STATIC  EQUILIBRIUM 

ii  ................................  12 

SECTION  IV.     COPLANAR  CONCURRENT  FORCES 

12.  Illustrative  Problem  ......................  14 

13.  General  Case  ...................    ......  17 

14.  Only  Two  Independent  Statements  or  Equations  .....    .....  19 

15.  Illustrative  Problem  ......................  IQ 

16.  Illustrative  Problem  ...........  •  ...........  20 

17.  Special  Case  ..........................  21 

18.  Conclusion  ................    .    .........  22 

19.  Problems  .................    '....'  ......  23 

SECTION  V.     COPLANAR  PARALLEL  FORCES 

20.  General  Considerations  .....................  24 

21.  Illustrative  Problems  ......................  27 

22.  Conclusion  ..........    ................  30 

SECTION  VI.     COPLANAR  NON-CONCURRENT  NON-PARALLEL  FORCES 

23.  Transformed  System  ......................  30 

24.  Illustrative  Problem  ......................  31 

25.  General  Case  .........................  32 

26.  Four  Groups  ..........................  33 

ix 


X  CONTENTS 

PAGE 

27.  Group  2 '.    „ 35 

28.  Groups ;'.-...' 36 

29 36 

30.  Group  4 36 

31 37 

32.  One  Unknown  Location 37 

33 38 

34.  Combinations 38 

35.  Illustrative  Problems 38 

36.  Problems 54 

CHAPTER  II 

SECTION  I.     SIMPLE  TRUSSES 

37.  Rigid  Body 60 

38.  A  Truss 61 

39.  Stability 61 

40.  The  Assumption 61 

41.  The  Application 62 

42.  The  External  Forces 62 

43.  The  Internal  Forces 62 

44.  The  Method  of  Joints 63 

45.  The  Method  of  Sections 68 

46.  Problems 71 

47 73 

SECTION  II.    STRUCTURES  REQUIRING  SPECIAL  CONSIDERATION 

48.  The  Fink  Truss .  73 

49.  The  Three-hinged  Arch 75 

50.  Illustrative  Problem 77 

51.  The  Cantilever  Bridge 80 

52.  Inclined  Loads '. 81 

SECTION  III.     BEAMS: — SHEAR  AND  BENDING  MOMENT  DIAGRAMS 

53.  A  Beam 83 

54.  The  Shear 83 

55.  The  Bending  Moment 83 

56.  The     Bending     Moment     Determined    from     the    Location-direction 

Diagram 84 

57.  Shear  and  Bending  Moment  Diagrams 86 

58.  Illustrative  Problems -,....'..  87 

59.  The  Structural  Engineer's  Parabola ....'..  91 

60.  Illustrative  Problem 95 

61.  Problems ' 97 

SECTION  IV.    FRAMES  HAVING  MEMBERS  WHICH  PERFORM  THE  FUNCTIONS  OF 

A  BEAM 

62 ^  ...... 99 

63.  Illustrative  Problem 09 

64.  Problems 102 

65.  The  Portal  Frame  .......,,. 102 


CONTENTS  XI 

PAGE 

66.  The  Portal  Frame 104 

67.  The  Portal  Frame 105 

68.  Problems 107 

CHAPTER  III 

ROOF  TRUSSES 

SECTION  I.     TYPES  AND  LOADS 

69.  Standard  Types 108 

70.  Purlins 109 

71.  The  Bracing •. "-. _.. no 

72.  Spacing  of  Trusses ...  •; in 

73.  The  Dead  Loads.    .....' in 

74.  The  Snow  Load 112 

75.  The  Wind  Load 113 

SECTION  II.     STRESS  ANALYSIS  FOR  WIND 

76.  Wall  Bearing  Trusses ,    ;  ;    .    .    .    .    . 115 

77.  Trusses  Supported  on  Columns .    .  116 

78.  Stress  Diagram  for  Wind  Unnecessary  ...,.«.. 124 

79.  Combined  Snow  and  Wind  Loads 125 

80.  Design  Stresses 125 

81.  Conclusions 126 

CHAPTER  IV 

BRIDGES 

SECTION  I.     STANDARD  TYPES 

82.  Standard  Types.      .   ."vv  »......«.. 128 

83.  Moving  Loads ...    .  ',  '.'..;*.   .    .    . 130 

SECTION  II.    DECK  BEAMS  AND  GIRDERS 

84.  Influence  Line  for  Reaction 130 

85.  Influence  Line  for  Shear  in  a  Beam 132 

86.  Influence  Line  for  Bending  Moment  in  a  Beam 132 

87.  Criterion  for  Maximum  Bending  Moment 134 

88.  Cooper's  Standard  Train  Loads 137 

89.  Illustrative  Problem 138 

90.  Problems 143 

91.  General  Criterion  for  Maximum  Bending  Moment 143 

92.  The  Point  of  Greatest  Maximum  Bending  Moment  in  a  Beam.    .    .    .  143 

93.  Problems 145 

94.  Criterion  for  Maximum  Shear  in  a  Beam 146 

95.  Problem 146 

SECTION  III.     PARALLEL  CHORD  TRUSSES 

96.  A  Railway  Truss  Bridge 146 

97.  Influence  Line  for  Maximum  Stress  in  a  Chord  Member.    ......  146 

98.  Illustrative  Problem  .    . 148 


Xll  CONTENTS 

PAGE 

99.  General  Criterion  for  Maximum  Chord  Stress 149 

100.  The  Stress  in  a  Web  Member  of  a  Parallel  Chord  Truss 151 

101.  Influence  Line  for  Shear  in  a  Panel 151 

102.  Criterion  for  Maximum  Shear  in  a  Panel  ....... 152 

103.  Illustrative  Problems.    .    .    ,   .C.  v 155 

104.  General  Criterion  for  Maximum  Shear  in  a  Panel 157 

105.  Impact 159 

106.  Stresses  in  a  2oo-ft.  Pratt  Truss 160 

107.  Counters 166 

108.  Truss  with  an  Odd  Number  of  Panels 167 

SECTION  IV.    PARKER  TRUSSES 

109.  Stress  in  Web  Member  of  Parker  Truss 169 

no.  Influence  Line  for  Web  Member  of  Parker  Truss 170 

in.  Criterion  for  Maximum  Stress  in  Web  Member  of  Parker  Truss  .    .    .  172 

112.  Illustrative  Problems 174 

113.  Problems 176 

114.  General  Criterion  for  Maximum  Stress  in  Web  Member  of  a  Parker 

Truss 177 

115.  Tension  in  a  Vertical  when  Counters  are  Used 182 

SECTION  V.    THE  BALTIMORE  TRUSS 

116.  The  Baltimore  Truss 184 

117.  LJ^e 184 

118.  U2U4 185 

119 189 

120 189 

121.  u2Ms 190 

122.    MsL4 190 

123 . J92 

124.  U^4 194 

125.  U2L2 197 

SECTION  VI.    THE  PENNSYLVANIA  TRUSS 

126.  Pennsylvania  Truss 197 

127.  U2U4 198 

128.  U2M3 ..^ 198 

129.  MaL4 199 

130.  U4L4 200 

SECTION  VII.    EQUIVALENT  UNIFORM  LOADS 

131 200 

132.  Examples 290 

133.  General  Considerations , 202 

134.  Triangular  Influence  Diagrams 203 

135.  Table  of  Equivalent  Uniform  Loads 205 


CONTENTS  Xlll 

CHAPTER  V 
DEFLECTION  OF  BEAMS 

SECTION  I.    THE  AREA-MOMENT  METHOD 

PAGE 

136.  General  Considerations 208 

137.  Method  of  Area-moments 208 

138.  First  Principle 211 

139.  Second  Principle t 212 

SECTION  II.     SIMPLE  BEAMS  OF  UNIFORM  CROSS-SECTION 

140.  The  Beam  in  Fig 212 

141.  Maximum  Deflection 215 

142.  Point  of  Maximum  Deflection 218 

143 220 

144.  Deflection  Under  Uniform  Load 221 

145.  Deflection  for  Load  of  Uniformly  Varying  Intensity 224 

146.  Deflection  for  Several  Concentrated  Loads 226 

147.  Deflection  for  Uniform  and  Concentrated  Loads 229 

SECTION  III.     MAXWELL'S  THEOREM  OF  RECIPROCAL  DISPLACEMENTS 

148.  Maxwell's  Theorem  of  Reciprocal  Displacements 231 

SECTION  IV.     CANTILEVER  BEAMS 
i49 -    • 232 

SECTION  V.  BEAMS  WITH  VARYING  CROSS-SECTION 

150.  General  Expressions 235 

151.  Beams  with  Varying  Depth 236 

152.  Beams  with  Cover  Plates 237 

CHAPTER  VI 

RESTRAINED  AND  CONTINUOUS  BEAMS 
SECTION  I.     RESTRAINED  OR  FIXED  BEAMS 

153.  General  Considerations 240 

154.  Restraint  at  One  End — Concentrated  Load 241 

155.  Restraint  at  One  End — Uniform  Load 242 

156.  Restraint  at  Both  Ends — Concentrated  Load 243 

157.  Restraint  at  Both  Ends — Uniform  Load 247 

SECTION  II.     CONTINUOUS  BEAMS 

158 248 

159.  Application  of  Maxwell's  Theorem 251 

160.  The  Conventional  Method 252 

161 256 

162.  The  General  Expressions 257 


XIV  CONTENTS 

PAGE 

J63 257 

164.  Two  Unequal  Spans,  Supporting  Unequal  Uniform  Loads.    .....  259 

165 260 

166 ....... 261 

167 263 

168.  Three  Equal  Spans — Uniform  Load 264 

169.  Four  Equal  Spans — Uniform  Load 266 

170.  Coefficients  for  Pier  Reactions *   .    .  267 

SECTION  III.     CLAPEYRON'S  THEOREM  or  THREE  MOMENTS 

171.  Clapeyron's  Theorem 268 

SECTION  IV.     PARTIALLY  CONTINUOUS  BEAMS 

172.  No  Shear  Transmitted 270 

173.  No  Moment  Transmitted 271 

SECTION  V.     CONTINUOUS  BEAMS  IN  FOUNDATIONS 

i?4 273 

175.  Case     I — Projections  Not  Limited  by  Site 275 

176.  Case    II — Projection  at  One  End  Limited  by  Site 277 

177.  Case  III — Both  Projections  Limited  by  Site 279 

CHAPTER  VII 

DEFLECTION  or  TRUSSES 

178.  Stress  and  Strain. 282 

SECTION  I.     ALGEBRAIC  METHOD 

179.  The  Algebraic  Solution 283 

180.  Illustrative  Problem 285 

SECTION  II.     GRAPHIC  METHOD 

181.  Willot  Diagrams 288 

SECTION  III.     GENERAL  CONSIDERATIONS 

182.  Temperature 297 

183.  Camber 298 

184.  Maxwell's  Theorem  of  Reciprocal  Deflection 299 

CHAPTER  VIII 

SWING  BRIDGES 
SECTION  I.     CENTER-BEARING  SWING  BRIDGES 

185.  General  Considerations *    .   « •  3O1 

186.  Conditions  of  Loading 3O1 

187.  Stresses  in  a  Swing  Span  .    . . 3°3 

188.  Positive  Shear  in  Panel  o-i 3°6 


CONTENTS  XV 

PAGE 

189.  Positive  Moment  about  U 3 — Case  III 310 

190.  Negative  Shear  in  Panel  o- 1 311 

191.  Shear  in  Panel  1-2 ,    .  •.':.    .    .    /-..    .    .    .  313 

192.  Shear  in  Panel  2-3 313 

193.  Shear  in  Panel  3-4 -.   ......    £'•*)  .    .  313 

194.  Shear  in  Panel  4-5 .    .    .    ...    .    .    .    .    .i.   -.    .    .    .    .  313 

195.  Shear  in  Panel  5-6 ...'......*,.....    .314 

196.  Moment  about  L2  .    .    .    .    .    .....   ',  •  ,    .    . 315 

197.  Moment  about  L4  .    .    .'  .    .    .  V.    ...  .- 316 

198.  Moment  about  U5 •.."-.    ...    .    .    .    . 317 

199.  Moment  about  Le 318 

200.  Case  V  Both  Arms  Loaded 318 

201.  Negative  Shear  in  Panel  5-6 * .  319 

202.  Moment  about  Le ; 319 

203.  Dead  Load:  Bridge  Open  ......................  320 

204.  Dead  Load:  Ends  Raised  .......    .    .    .    .  , 320 

205.  Combinations ..    ...    .    ,    . 320 

206.  Reactions  from  Willot  Diagram 320 

SECTION  II.     RIM-BEARING  SWING  BRIDGES 

207 V  .    ...    ...   323 

208.  Partial  Continuity 325 

INDEX 327 


ESSENTIALS 

IN  THE 

THEORY  OF   FRAMED  STRUCTURES 

CHAPTER  I 

EQUILIBRIUM  OF  COPLANAR  FORCES 
SEC.  I.  INTRODUCTION 

i.  Mechanics. — The  theory  of  stresses  in  framed  structures 
is  the  development  of  a  few,  simple,  fundamental  principles  of 
mechanics.  The  science  of  mechanics  is  a  study  of  motion. 
The  idea  of  motion  is  closely  related  to  the  ideas  of  space,  time 
and  mass. 

We  study  first  the  motion  of  a  body  without  regard  to  the 
time  consumed  in  the  motion,  or  the  mass  of  the  thing  moved. 
This  part  of  mechanics  is  called  "  Geometry  of  Motion." 

Secondly,  the  introduction  of  the  idea  of  time  leads  us  to  con- 
sider velocity  and  acceleration.  This  study  is  called  "Kine- 
matics." 

Finally,  when  the  body  is  assigned  to  a  certain  mass,  we 
consider  the  ideas  of  energy,  momentum  and  force.  This  part 
of  mechanics  is  called  "  Dynamics"  and  is  divided  into  two 
branches— " Kinetics"  and  "Statics." 

Kinetics  treats  in  the  most  general  way  of  changes  in  motion 
produced  by  forces.  Statics  is  a  consideration  of  those  cases 
in  which  no  change  of  motion  is  produced,  and  the  element  of 
time  is  not  a  factor. 

The  fundamental  conception  of  stress  analysis  in  framed 
structures  is  acquired  by  a  systematic  study  of  that  part  of 
Statics  which  treats  of  the  "Equilibrium  of  Coplanar  Forces." 


2  THEORY    OF   FRAMED    STRUCTURES  CHAP.  I 

This  topic  is  not  a  new  one  to  the  student  of  Physics  and 
Mechanics.  In  both  of  these  courses  it  is  treated  in  a  general 
way  in  connection  with  many  kindred  topics,  equally  important 
in  other  fields  of  engineering  science. 

The  first  chapter  of  this  book  has  been  prepared  for  the 
purpose  of  making  an  extended  investigation  of  the  "  Equilib- 
rium of  Coplanar  Forces,"  to  the  end  that  the  reader  may 
acquire  a  more  specific  knowledge  of  the  subject  and  the  power 
to  apply  this  knowledge  with  reason  and  common  sense.  It 
is  the  most  important  chapter  in  the  book;  it  forms  the  base 
upon  which  rest  the  principles  of  each  succeeding  chapter,  and 
should  be  thoroughly  mastered. 

2.  Methods  and  Mental  Attitude. — Whenever  we  engage  in  a 
new  piece  of  work,  it  is  wise  to  consider  the  tools  with  which  we 
may  accomplish  our  task,  and  the  manner  in  which  these  tools 
can  be  used  most  advantageously.  The  man  who  is  about  to 
cut  down  a  tree  decides,  not  only  whether  he  will  use  an  axe  or  a 
saw,  but  also  (and  quite  unconsciously,  perhaps),  whether  he 
will  operate  the  chosen  implement  with  his  hands  or  his  feet. 

The  student  will  do  well  to  emulate  the  prudence  of  the  wood 
cutter  and,  for  the  moment,  consider  the  tools  or  methods  by 
which  stresses  may  be  determined.  He  should  also  take  thought 
of  the  much  more  important  question — By  what  mental  pro- 
cesses can  these  tools  most  successfully  be  operated? 

There  are  two  methods  by  which  stresses  may  be  determined : 
(a)  the  algebraic  method,  which  makes  use  of  numerals  or 
symbols;  (b)  the  graphic  method,  which  represents  quantities 
by  the  length  and  direction  of  lines. 

Graphic  methods  are  not  limited  to  the  solution  of  static 
problems.  The  operations  of  addition,  subtraction,  multipli- 
cation, division,  involution  and  evolution  can  be  performed 
graphically;  and  many  problems  in  calculus  are  more  readily 
comprehended  when  illustrated  by  the  graphic  method. 
Whether  the  algebraic  or  the  graphic  method  is  better,  is  a 
question  which  depends  for  its  answer  upon  the  individual 
problem.  However,  if  each  problem  is  solved  by  both  methods, 
the  fundamental  principles  are  more  clearly  set  before  the 
beginner. 


SEC.  I  EQUILIBRIUM    OF    COPLANAR   FORCES  3 

The  student  should  always  remember  that  mental  develop- 
ment is  more  to  be  desired  than  the  mere  accumulation  of 
information,  which  is  of  little  value  unless  its  possessor  has  the 
mental  power  or  capacity  for  using  it  intelligently.  Since  the 
student  can  acquire  intellectual  power  only  by  intellectual 
exercise,  he  is  frequently  asked  by  his  instructor  to  think. 
This  request  may  be  interpreted  in  several  different  ways.  One 
may  literally  comply  with  the  request  if  he  memorizes,  imagines, 
reasons  or  performs  one  of  several  other  minor  intellectual  funct- 
ions. The  intellectual  development,  however,  varies  with  the 
form  of  exercise — a  fact  which  the  instructor  should  not  fail 
to  recognize.  Although  the  training  of  the  memory  is  an 
important  intellectual  process,  imagination  and  reasoning 
should  not  be  neglected.  Which  of  these  three  plays  the  lead- 
ing role  in  the  drama  of  life  is  a  question  which  is  answered 
differently  perhaps  by  the  linguist,  poet  and  scientist.  It  is  a 
significant  fact,  however,  that  the  majority  of  people  who 
plead  guilty  to  possessing  a  poor  memory  are  much  offended 
when  accused  of  having  little  imagination  or  inferior  judgment. 

The  degree  to  which  the  student  will  exercise  his  memory,  his 
imagination  or  his  reason  in  the  preparation  of  a  lesson,  depends 
to  a  considerable  degree  upon  the  manner  in  which  the  subject 
is  presented  in  the  textbook.  If  he  sees  a  formula,  his  first  im- 
pulse is  to  memorize  it,  too  often  in  parrot-like  form;  while  its 
history,  meaning  and  limitations  are  to  him  a  matter  for  second- 
ary consideration.  In  a  text  book  or  in  the  class-room,  formulas 
should  be  given  what  little  place  they  deserve  at  the  end,  and 
not  at  the  beginning  of  a  subject;  for  the  simple  reason  that 
algebraic  manipulations  obscure  the  physical  relations  which 
should  be  visualized  to  the  highest  degree. 

It  may  appear  to  the  casual  reader  that  imagination  is  an 
intellectual  process  exercised  exclusively  by  the  artist  or  poet. 
Such  is  not  the  case.  A  structure,  whether  it  be  a  steam  engine, 
electric  generator  or  a  railway  bridge  must  be  visualized  first 
in  the  mind  of  the  engineer  before  it  can  become  a  reality;  or 
even  before  it  can  take  shape  and  size  upon  the  designing 
board. 

In  preparing  the  manuscript  for  this  book,  the  author  has 


4  THEORY    OF   FRAMED    STRUCTURES  CHAP.  I 

endeavored  to  present  each  topic  in  such  form  that  the  student 
will  find  employment  for  his  imagination  and  reason  as  well  as 
for  his  memory.  Incidentally,  he  will  acquire  a  working  knowl- 
edge of  the  theory  of  stresses  in  framed  structures. 

3.  Forces. — The  word  "force"  is  used  to  express  a  variety  of 
ideas.  We  speak  of  physical,  mental  and  moral  forces,  etc. 
But  for  scientific  purposes  it  is  necessary  to  have  a  single  definite 
meaning.  In  Physics  and  Mechanics  we  define  force  as  the 
cause  of  acceleration  or  change  in  velocity;  i.e. 

force  =  rate  of  change  of  momentum 

and  momentum  =  mass  X  velocity 

hence  force  =  mass  X  rate  of  change  of  velocity 

or  since          rate  of  change  of  velocity  =  acceleration 
therefore  force  =  mass  X  acceleration. 

All  bodies  are  constantly  acted  upon  by  the  force  of  gravita- 
tion; and,  except  in  the  special  case  of  bodies  falling  in  a  vacuum, 
by  other  forces  also.  A  moving  train  on  a  straight,  level  track 
is  usually  subject  to  the  action  of  four  forces:  the  action  of 
gravitation  "pulling  down"  and  the  reaction  of  the  rails  "push- 
ing up";  the  "pull"  of  the  engine  in  one  direction  and  the  retard- 
ing action  of  friction  and  atmosphere  in  the  opposite  direction. 

If  the  reaction  of  the  rails  equals  the  action  of  gravitation, 
the  train  will  remain  at  the  same  level.  If  the  engine  pull  is 
greater  than  the  retarding  action,  the  difference  represents  the 
force  which  accelerates  the  speed  of  the  train;  thereby  increas- 
ing its  velocity  and  momentum.  On  the  other  hand,  if  the 
force  represented  by  the  engine  pull  equals  the  retarding  forces, 
the  train  is  moving  at  a  constant  speed,  and  there  is  no  rate  of 
change  of  velocity  or  momentum.  The  two  equal  and  opposite 
forces  "balance"  and  their  algebraic  sum  is  zero.  Thus  if  a 
body  is  at  rest,  or  is  moving  in  a  straight  line  with  constant 
velocity,  the  acceleration  is  zero,  there  is  no  change  in  momen- 
tum, and  the  body  is  said  to  be  in  a  state  of  equilibrium. 

The  theory  of  stresses  in  framed  structures  is  developed  by  a 
study  of  the  action  of  forces  on  bodies  at  rest.  We  need  give 
no  special  attention  to  velocity,  momentum  or  acceleration  in 
our  conception  of  the  idea  of  force.  For  our  present  discussion 
we  may  therefore  properly  conceive  of  a  force  as  the  idea  ex- 
pressed in  the  words  "push"  and  "pull." 


SEC.  I  EQUILIBRIUM   OF   COPLANAR  FORCES  5 

4.  The  Three  Elements  of  a  Force. — We  know  that  the  effect 
of  a  force  of  50  Ib.  pushing  vertically  (downward)  upon  the  head 
is  quite  different  from  the  effect  of  a  force  of  80  Ib.  pulling  hori- 
zontally at  the  feet;  and  thus  from  our  personal  experiences  we 
may  observe  that  the  characteristics  or  elements  of  a  force  are: 

1.  Magnitude. 

2.  Direction. 

3.  Location. 

The  effect  of  the  action  of  a  force  upon  a  body  cannot  be 
determined  until  all  three  elements  are  known. 

The  magnitude  or  intensity  of  a  force  is  usually  expressed  in 
pounds,  units  of  1,000  Ib.,  or  tons. 

The  direction  may  be  indicated  by  the  angle  between  the  hori- 
zontal and  the  line  along  which  the  force  acts.  For  our  purpose, 
however,  it  will  be  found  more  advantageous  to  designate  the 
direction  by  giving  the  slope  or  bevel  of  the  line  along  which  the 
force  acts. 

The  location  of  a  force  is  known  when  the  position  (with  refer- 
ence to  the  body)  of  any  point  in  the  line  representing  the  direc- 
tion is  given. 

The  sense  in  which  a  force  acts  along  a  given  direction,  i.e., 
up  or  down,  toward  the  right  or  left,  is  represented  graphically 
by  an  arrow-head.  In  an  algebraic  solution  the  sense  is  desig- 
nated by  the  signs  +  or  —  in  connection  with  the  magnitude 
of  the  force.  The  sense  of  a  force  is  not  ranked  as  an  element, 
since  it  requires  no  separate  equation  for  its  determination. 

5.  Definitions. — Forces  are  coplanar  when  acting  in  the 
same  plane;  non-coplanar,  when  acting  in  different  planes. 

Forces  are  concurrent  when  their  directions  meet  in  a  point, 
and  non-concurrent  when  their  directions  do  not  so  meet. 

Several  forces  under  consideration  are  called  a  system  of 
forces. 

A  couple  is  a  system  of  two  parallel  forces,  equal  in  magnitude, 
but  opposite  in  sense. 

Two  systems  of  forces  are  equivalent  if  one  system  may  be 
substituted  for  the  other  without  changing  the  state  of  rest  or 
motion  of  the  body  upon  which  they  act. 


O  THEORY   OF   FRAMED    STRUCTURES  CHAP.  I 

The  resolution  of  forces  is  the  process  of  resolving  the  forces 
of  one  system  into  an  equivalent  system  having  a  greater 
number  of  forces. 

The  composition  of  forces  is  the  process  of  combining  or  re- 
ducing the  forces  of  one  system  into  an  equivalent  system  hav- 
ing fewer  forces.  If  the  equivalent  system  reduces  to  a  single 
force,  that  force  is  called  the  resultant  of  the  system;  and  each 
force  in  the  system  is  a  component  of  the  resultant  force.  All 
systems,  however,  cannot  be  reduced  to  a  single  force;  but  all 
coplanar  systems  not  in  equilibrium  can  be  reduced  to  either  a 
single  force  or  a  couple. 

SEC.  II.     RESOLUTION  AND  COMBINATION  OF  FORCES 

6.  Parallelogram  of  Forces.— The  law  of "  Triangle  of  Forces" 
seems  to  have  been  discovered  in  the  year  1608  by  Simon  Stevin, 
a  Flemish  military  engineer,  who  conceived  the  idea  of  the  reso- 
lution and  composition  of  forces,  by  his  investigation  of  the 
properties  of  the  inclined  plane.  He  derived  the  law  from 
experimental  data.  Mathematical  proofs  of  this  law  have 
frequently  been  presented;  but  several  eminent  writers  of  the 
present  day  seem  to  be  content  with  the  statement  that  the  law 
is  so  fundamental  that  it  cannot  be  deduced  from  anything  more 
simple  than  itself. 

An  experimental  demonstration  of  the  law  is  illustrated  in 
Fig.  i.  Three  smoothly  running  pulleys  are  mounted  on  a 
vertical  drawing  board  at  any  points  A,  B,  and  C  not  in  the 
same  straight  line.  Three  strings  are  tied  together,  placed 
over  the  pulleys  and  made  to  support  three  unequal  weights, 
P,  Q,  and  S.  Under  the  action  of  gravitation,  the  weights  will 
adjust  themselves  and  come  to  a  state  of  rest.  Mark  on  the 
board  the  position  of  the  knot  at  o.  Add  a  small  weight  to 
P  and  note  the  change  in  the  position  of  the  knot.  Remove  the 
small  weight  and  observe  that  the  knot  returns  to  its  original 
position.  Do  likewise  with  Q,  and  with  S.  Now  increase 
each  weight  by  an  equal  amount  and  note  that  the  knot  changes 
its  position.  Increase  each  weight  by  one-half  or  one-third  of 
itself  and  observe  that  the  position  of  the  knot  is  not  changed. 


SEC.  II  EQUILIBRIUM    OF    COPLANAR   FORCES  7 

Under  the  action  of  gravitation,  each  weight  produces  a 
corresponding  tension  in  the  string  which  supports  it;  and  we 
have  a  system  of  three  concurrent  forces  in  equilibrium. 

In  order  that  a  line  may  be  employed  to  represent  a  force  it 
is  necessary  that:  (i)  its  length  should  indicate  the  magnitude 
of  the  force;  (2)  its  direction,  with  an  arrow-head  to  show  the 
sense,  should  correspond  to  the  line  of  action  of  the  force; 
(3)  the  location  of  a  point  in  the  line  should  be  given. 

The  lines  oA ,  oB  and  oC  with  arrow-heads  to  show  the  sense, 
represent  the  directions  of  the  three  forces,  and  the  point  o 
designates  their  location.  By  choosing  a  convenient  scale- 
ratio,  let  us  say,  10  Ib.  per  inch,  the  lengths  oa,  ob  and  oc  may  be 
laid  off  on  the  lines  oA ,  oB  and  oC  to  represent  the  magnitudes, 


FIG.  i. 

P,  Q  and  S  respectively.  The  sequence  of  the  letters  which 
designate  the  force  may  be  used  to  indicate  the  sense  of  the 
force.  Thus  oa  signifies  that  the  sense  of  the  force  P  is  upward 
from  o  to  a,  while  ao  would  represent  a  force  the  equivalent  of  P 
in  all  respects  except  that  the  sense  would  be  opposite,  i.e., 
downward  from  a  to  o.  The  forces  P,  Q  and  S  are  thus  com- 
pletely represented  by  the  lines  oa,  ob  and  oc,  since  each  line 
designates  the  magnitude,  direction  (including  sense)  and 
location  of  its  respective  force. 

Draw  a  line  through  a,  parallel  to  ob,  and  a  line  through  b, 
parallel  to  oa  completing  the  parallelogram  oadb.     Measure  the 


8  THEORY    OF   FRAMED    STRUCTURES  CHAP.  I 

length  of  the  diagonal  od,  with  the  same  scale  as  before.  Ob- 
serve that  the  length  do  equals  the  length  oc;  and  that  the  points 
Cj  o  and  d  are  in  a  straight  line. 

We  say  that  the  force  5,  represented  by  do,  is  the  equilibrant 
of  the  forces  P  and  Q,  since  it  holds  them  in  equilibrium  or 
balances  them.  With  equal  propriety  we  can  say  that  a 
force  Rj  represented  by  od  having  equal  magnitude,  the  same 
direction  and  location  as  the  force  5  but  opposite  sense,  would 
hold  the  force  5  in  equilibrium,  if  the  forces  P  and  Q  were 
removed.  Hence,  the  force  R  is  called  the  resultant  of  the 
forces  P  and  Q,  since  it  may  be  substituted  for  them  without 
disturbing  the  state  of  equilibrium.  The  law  of  the  parallelo- 
gram of  forces  may  be  stated  as  follows : 

If  two  concurrent  coplanar  forces  are  represented  in  magni- 
tude and  direction  by  the  adjacent  sides  of  a  parallelogram,  so 
constructed  that  the  sense  of  each  force  is  the  same  with  respect 
to  their  point  of  intersection;  then  the  resultant  of  the  two 
forces  is  represented  in  magnitude  and  direction  by  the  diagonal 
of  the  parallelogram  through  the  point  of  intersection.  If  the 
sense  of  the  two  forces  is  away  from  the  point  of  intersection, 
the  sense  of  the  resultant  is  away  from  the  point  and  vice  versa. 

7.  Triangle  of  Forces. — Since  a  diagonal  divides  a  parallelo- 
gram into  two  equal  and  similar  triangles,  it  is  necessary  to 
construct  but  one  of  the  triangles  in  order  to  find  the  resultant. 
Instead  of  representing  the  two  forces  P  and  Q  by  the  lines  oa  and 
ob  drawn  from  o,  we  may  draw  oa  representing  the  forceP;  and 
from  a  draw  ad  representing  the  force  Q.  Or,  we  may  first  draw 
ob  to  represent  the  force  Q,  and  from  b  draw  bd  representing  the 
force  P.  In  either  construction  the  line  od  represents  the 
resultant  R,  of  the  forces  P  and  Q;  while  the  line  do  represents 
their  equilibrant  S. 

In  a  concurrent  system,  all  forces  act  through  a  common  point 
which  establishes  the  location  of  each  force,  and  we  shall  be 
concerned  chiefly  with  the  determination  of  unknown  magni- 
tudes and  directions.  For  this  reason  it  is  not  essential  that 
the  triangle  be  constructed  in  connection  with  the  point  of 
concurrency  o.  The  triangle  efgj  for  example,  having  sides 
respectively  parallel  to  ob,  bd  and  do  will  serve  our  purpose  in  a 


SEC.  II 


EQUILIBRIUM   OF   COPLANAR   FORCES 


consideration  of  the  forces  P,  Q  and  5,  equally  as  well  as  the 
similar  triangle  obd.  The  homologous  sides  of  the  two  triangles 
are  proportional  and  parallel — an  important  fact  to  remember. 
The  triangles  differ  in  size  simply  because  different  scale-ratios 
were  used  in  their  construction. 

If  the  sides  of  a  triangle  represent  three  forces  in  equilibrium, 
the  arrow-heads  will  indicate  a  continuous  course  around  the 
perimeter  of  the  triangle.  If,  however,  the  three  sides  of  the 
triangle  represent  two  forces  and  their  resultant,  the  arrow- 
head of  the  resultant  will  be  opposed  to  the  arrow-heads  of  the 
two  other  forces. 

The  law  of  the  triangle 
of  forces  may  be  stated  as 
follows:  Three  concurrent 

co planar  forces  in  equilib-  \^ -^      s  ^ 

rium  may  be  represented  in 
magnitude,  direction  and 
sense  by  the  three  sides  of 
a  triangle  taken  in  contin- 
uous order.  Any  one  of 
the  three  forces  may  be 
made  to  represent  the  resul- 
tant of  the  other  two  by 
reversing  its  sense. 

8.  Magnitude-direction 
Diagram. — When  the  re- 
sultant of  several  concurrent  forces  is  required,  the  resultant  of 
any  two  may  be  found  from  the  force  triangle;  and  this  resultant 
combined  with  a  third  force  to  obtain  the  resultant  of  the  three 
forces.  This  process  may  be  continued  until  all  forces  have 
been  combined.  A  system  of  four  coplanar,  concurrent  forces 
P,  Q,  S  and  T  intersecting  at  O  is  represented  in  Fig.  20.  In 
Fig.  2b,  ac  is  the  resultant  of  P  and  Q;  ad  is  the  resultant  of  ac 
and  S;  and  finally  ae,  the  resultant  of  ad  and  T,  is  the  resultant 
R  of  the  entire  system.  Evidently  the  force  £,  represented 
by  ea,  having  the  same  magnitude  and  direction  as  R  but 
opposite  in  sense,  is  the  equilibrant  which,  if  applied  at  0 
would  secure  equilibrium.  The  partial  resultants  ac  and  ad 


FIG.Sb 


10  THEORY  OF  FRAMED  STRUCTURES        CHAP.  I 

are  obviously  superfluous,  since  the  figure  abode  could  have 
been  drawn  without  them. 

Textbooks  in  Physics  and  Mechanics  usually  refer  to  the 
diagram  in  Fig.  26,  as  a  force  polygon.  The  name  is  misleading. 
The  figure  does  not  completely  represent  a  force — it  represents 
only  two  of  the  three  essential  elements.  Magnitude-direction 
diagram  is  a  more  distinctive  term.  The  order  in  which  the 
forces  are  combined  is  of  no  particular  consequence.  In 
Fig.  20  the  forces  have  been  combined  in  the  order,  (),  T,  5,  P ; 
but  the  resultant  R,  or  the  equilibrant  E,  are  the  same  as  in 
Fig.  26. 

The  known  magnitudes  and  directions  were  laid  off  from  a  to 
e\  and  the  magnitude  and  direction  of  the  equilibrant  E  was 
found  by  drawing  the  line  ea,  which  necessarily  closed  the 
figure.  It  follows  that  a  graphic  solution  for  equilibrium  must 
comply  with  the  following  law: 

If  a  system  of  coplanar,  concurrent  forces  is  in  equilibrium, 
the  magnitude-direction  diagram  will  close,  and  the  sense  will  be 
continuous. 

Of  course  the  magnitude-direction  diagram  could  have  been 
closed  by  any  series  of  straight,  broken  lines,  continuous  from 
e  to  a;  but  certain  limitations  must  be  imposed  upon  the  number 
of  unknown  magnitudes  and  directions  if  a  single  solution  is  to 
be  realized.  We  shall  refer  to  this  in  detail  in  connection  with 
the  algebraic  solution. 

9.  Rectangular  Components. — A  force  may  be  resolved  into 
two  components  by  applying  the  triangle  law  conversely.  Since 
an  infinite  number  of  triangles  can  be  drawn  having  one  side  in 
common,  it  follows  that  the  problem  of  finding  the  two  com- 
ponents of  a  given  force  is  indeterminate  unless  certain  conditions 
are  imposed  upon  the  components.  Rectangular  components 
have  directions  intersecting  at  an  angle  of  90°,  and  are  very 
helpful  in  algebraic  solutions;  since  an  inclined  force  can  thus 
be  resolved  into  and  replaced  by  its  horizontal  and  vertical 
components.  Thus  in  Fig.  i  the  three  forces  in  equilibrium, 
P,  Q  and  S  are  resolved  into  horizontal  and  vertical  components 
by  the  right  triangles  fkg,  ejf  and  gie  respectively.  The  hori- 
zontal components  ej  and  fk  are  balanced  by  the  component 


SEC.  II 


EQUILIBRIUM    OF    COPLANAR   FORCES 


II 


gi,  being  equal  in  magnitude  and  opposite  in  sense.  Likewise 
the  vertical  components  jf  and  kg  are  balanced  by  the  com- 
ponent ie. 

Whenever  the  magnitude-direction  diagram  is  a  closed  figure, 
whether  a  triangle,  as  efg  (Fig.  i);  or  any  polygon,  as  abode 
(Fig.  2) ,  the  horizontal  components  will  balance  and  the  vertical 
components  will  balance.  If  opposite  signs  (+  and  — )  are 
given  to  magnitudes  of  opposite  sense,  the  algebraic  sum1  of  the 
horizontal  components  is : 


and  likewise 


=  o 

ZF  =  o 


These  two  equations  in  an  algebraic  solution  correspond  to  the 
closing  of  the  magnitude-direction  diagram  in  a  graphic  solution. 

10.  Moment  of  a  Force.— The  product  of  the  magnitude  of  a 
force  and  its  perpendicular  distance  from  a  point  is  called  the 
moment  of  the  force  about  the 
point.  The  perpendicular  dis- 
tance is  called  the  arm  of  the  force, 
and  the  point  is  known  as  the  cen- 
ter of  moments.  If  the  magnitude 
of  a  force  P  is  8  lb.,  and  its  arm 
from  a  point  0  is  3  ft.,  the  mo- 
ment of  the  force  P  about  the  point 
Ois24ft.-lb. 

In  1687,  °r  about  80  years  after  Stevin  had  published  his 
demonstration  of  the  triangles  of  forces,  Pierre  Varignon  pre- 
sented before  the  Paris  Academy,  the  "Principle  of  Moments." 
This  principle  is  known  as  Varignon's  Theorem.  He  stated 
that  "the  moment  of  the  diagonal  of  a  parallelogram  of  forces 
equals  the  sum  of  the  moments  of  the  other  two  sides."  His 
proof  is  somewhat  as  follows: 

The  parallelogram  CADB  (Fig.  3)  represents  the  forces  P  and 
Q  and  their  resultant  R.  The  lengths  of  the  perpendiculars 
from  the  center  of  moments  O  to  the  lines  representing  the 
forces  Pj  Q  and  R,  are  p,  q  and  r  respectively: 

JThe  symbol  S  represents  the  idea  expressed  in  the  words  "algebraic  sum." 


12  THEORY    OF    FRAMED    STRUCTURES  CHAP.  I 

area   AOCD  =  area  AOCA  +  area  AOAD  +  area  AACD 

lrR  =  -2pP  +  lsQ  +  ±tQ 
hence,  rR  =  pP  +  g<2 

Varignon  showed  that  the  center  of  moments  0  may  be 
located  outside  the  parallelogram,  within  it,  or  on  one  of  its 
sides. 

It  is  obvious  that  the  moment  rR  of  the  resultant  Rj  acting 
counter-clockwise  about  the  point  0,  may  be  balanced  by  the 
moment  rE  of  the  equilibrant  E  acting  clockwise.  If  opposite 
signs  are  given  to  the  moments  of  the  two  forces  according  as 
they  act  clockwise  or  counter-clockwise,  their  algebraic  sum  is 
zero,  or 

rR  -  rE  =  o 
therefore,  pP  +  qQ  -  rE  =  o 

P,  Q  and  E  represent  three  coplanar,  concurrent  forces  in 
equilibrium;  but  the  principles  may  be  generalized  so  as  to 
include  any  number  of  concurrent  forces  by  combining  them 
into  partial  resultants  after  the  manner  of  Fig.  2.  Hence,  if  a 
system  of  coplanar,  concurrent  forces  is  in  equilibrium,  the 
algebraic  sum  of  the  moments  of  all  the  forces  about  any  point 
in  their  plane  is 

I.M  =  o 

SEC.  III.     THE  THREE  FUNDAMENTAL  STATEMENTS  OR  EQUA- 
TIONS OF  STATIC  EQUILIBRIUM 

II.  Problems  in  stress  analysis  deal  in  general  with  the  action 
of  coplanar  forces  upon  bodies  at  rest,  and  the  solution  of  any 
problem  consists  in  finding  the  unknown  elements  of  these 
forces.  If  the  problem  is  "  statically  determinate,"  the  num- 
ber of  unknown  elements  will  not  exceed  the  number  of  inde- 
pendent statements  or  equations  which  can  be  written  concern- 
ing the  static  equilibrium  of  the  body  upon  which  the  forces  are 
acting.  It  is  a  simple  matter  to  write  these  equations  if  we 
consider  carefully  the  conditions  under  which  a  body  may  be  at 
rest. 


SEC.  Ill  EQUILIBRIUM    OF    COPLANAR   FORCES  13 

If  a  body  is  at  rest  under  the  action  of  any  system  of  forces 
lying  in  a  vertical  plane,  the  following  statements  are  self- 
evident  from  the  very  nature  of  the  case. 

1.  The  body  is  not  moving  toward  the  right  or  toward  the 
left,  consequently  the  horizontal  magnitudes  acting  toward  the 
right  are  balanced  by  the  horizontal  magnitudes  acting  toward 
the  left.     If  the  horizontal  magnitudes  acting  toward  the  right 
are  given  a  positive  sign,  and  the  horizontal  magnitudes  acting 
toward  the  left  are  given  a  negative  sign,  the  algebraic  sum  of 
the  horizontal  magnitude  is 

I#  =  o  (i) 

2.  The  body  is   not  moving  upward  or  downward,  conse- 
quently the  vertical  magnitudes  acting  upward  are  balanced 
by  the  vertical  magnitudes  acting  downward.     If  the  vertical 
magnitudes  acting  upward  are  given  a  positive  sign,  and  the 
vertical   magnitudes    acting  downward  are  given  a  negative 
sign,  the  algebraic  sum  of  the  vertical  magnitude  is 

IF  =  o  (2) 

3.  The  body  is  not  rotating,   either  clockwise  or  counter- 
clockwise, about  any  point  in  the  plane  of  the  forces;  conse- 
quently the  moments  of  all  the  magnitudes  acting  clockwise 
about  any  point  in  the  plane  of  the  forces  are  balanced  by  the 
moments  of  all  the  magnitudes  acting  counter-clockwise  about 
the  same  point.     If  clockwise  moments  are  given  a  positive 
sign,   and   counter-clockwise  moments   are   given   a   negative 
sign,  the  algebraic  sum  of  the  moments  about  any  point  in  the 
plane  of  the  forces  is 

ZM  =  (3) 

These  three  statements  or  equations  are  the  conditions  which 
must  be  fulfilled  by  any  system  of  forces  acting  in  a  vertical 
plane  upon  a  body  at  rest,  whether  the  system  be  concurrent, 
parallel  or  non-concurrent  non-parallel.  Equations  (i),  (2) 
and  (3)  embody  the  three  fundamental  principles  of  static 
equilibrium  for  coplanar  forces.  They  are  as  fundamental  and 
important  as  they  are  simple. 

Equations  (i)  and  (2)  are  easily  written  for  any  system,  after 
each  inclined  force  has  been  resolved  and  replaced  by  its  hori- 


14  THEORY  OF  FRAMED  STRUCTURES        CHAP.  I 

zontal  and  vertical  components.  Instead  of  resolving  the 
forces  along  horizontal  and  vertical  axes,  we  may  choose  any 
other  pair  of  axes  inclined  at  any  angle  and  equate  to  zero  the 
algebraic  sum  of  the  components  parallel  to  each  axis.  Thus, 
we  may  write  many  equations  of  types  (i)  and  (2).  We  may 
write  also  many  equations  of  type  (3)  by  choosing  different 
points  for  the  center  of  moments. 

Since  a  single  solution  is  possible  only  when  the  number  of 
unknown  elements  does  not  exceed  the  number  of  independent 
equations  which  may  be  written,  the  question  immediately 
arises — How  many  equations,  written  as  above,  will  be  inde- 
pendent for  a  given  system  of  forces?  In  answering  this  ques- 
tion we  must  distinguish  between  concurrent,  parallel,  and 
non-concurrent  non-parallel  systems;  and  for  this  reason  the 
three  systems  will  be  considered  separately. 

SEC.  IV.  COPLANAR  CONCURRENT  FORCES 

12.  Illustrative  Problem. — Five  concurrent  forces,  A,  B,  C, 
D  and  E  are  located  by  the  point  0  in  Fig.  4#.  The  direction  of 
each  force  is  given,  and  the  magnitudes  of  D  and  E  are  known. 
We  shall  attempt  to  find  (a)  by  the  algebraic  method  and  (b) 
by  the  graphic  method,  the  magnitudes  of  A ,  B  and  C  necessary 
for  equilibrium. 

(a)  Algebraic  Method. — We  first  resolve  each  inclined  force 
into  horizontal  and  vertical  components,  and  indicate  them  in 
the  sketch  (Fig.  46). 

The  magnitude  of  the  force  D  is  100  Ib.  It  acts  to  the  left 
and  upward,  sloping  at  the  bevel  of  3  in.  horizontally  to  4  in. 
vertically;  as  indicated  by  the  arrow-head  at  D  and  the  sides 
fj  and  jk  of  the  triangle  $&.  Hence, 

fj  :jk  :fk  1:3  -.4  15 
Let  D  =  100  Ib.  represent  the  magnitude  of  the  force, 

Dh  represent  the  horizontal  component, 
and  Dv  represent  the  vertical  component. 
Then,  if  fjk  is  considered  as  a  force  triangle, 
Dh:Dv:D::fj  :jk  :fk 


SEC.     IV 
or 


EQUILIBRIUM  OF  COPLANAR  FORCES 


Dh  :  Dv  :  100  ::  3  in.  14  in.  :  5  in. 
Dh  =  60  lb.,  acting  to  the  left 
and  Dv  =  80  lb.,  acting  upward. 

The  magnitude  and  sense  of  the  force  4  are  unknown.  For 
the  present  we  shall  assume  (or  guess)  that  the  sense  is  away 
from  the  point  0,  and  let 

Ah  =  8a.,  acting  to  the  right, 
Av  =  150.,  acting  upward; 


then 
and 


A  =  a\/82  +  is2  =  170. 


60       £4 

0  8%    /*£ 

5 
? 

P*2" 

5b 

FIGAb 

_    t 

FIG.4e 


Assume  that  the  force  B  acts  away  from  0,  and  let 
Bh  =  i2b,  acting  to  the  right, 
Bv  =  56,  acting  downward, 

and  B  =  6\/i22  +  52  =  136. 


1  6  THEORY  OF  FRAMED  STRUCTURES        CHAP.  I 

The  forces  E  and  C  (C  is  assumed  to  act  upward)  and  the 
components  of  the  inclined  forces  D,  A  and  B  are  shown  in  the 
sketch  (Fig.  46).  Three  independent  equations  are  required 
for  a  solution  of  the  three  unknown  quantities  C,  a  and  b. 


or  forces  acting  to  the  left  =  forces  acting  to  the  right. 

60  =  64  +  80  +  126  (i) 

IF  =  o 

or   forces   acting   upward  =  forces  acting  downward. 

C  +  80  +  150  =  56  (2) 

2M  =  o 

or  clockwise  moments  =  counter-clockwise  moments 
about  any  point.    Let  us  choose  the  point  F.     The  arm  of  the 
horizontal  forces  is  p  =  2  in.  and  the  arm  of  the  vertical  forces 
is  q  =  3  in. 

IM 

O  O 

Then   2  X  64  =  128  2  X  60  =  120 

2  X  80  =  160  3  X   C  =   $C 

2  X  126  =  246  3  X  80  =  240 

3  X  56  =  156  3  X  150  = 


128+160+    sgb  =                   36o+    3^  +  45<*       (3) 
Equations  (i),  (2)  and  (3)  may  be  reduced  to 

80  +  126  =  —4  (10) 

i5<*  -  5&    +    C  =  -80  (20) 

290  —  396  +  3C  =  —232  (30) 

Multiply  (20)  by  3  and  subtract  (30)  therefrom 

450  —  156  +  3C  =  —240 

290  -  396  +  3C  =  -232 

i6a  +  246  =  —  8 

or  80+126  =  —4  (4) 

Equations  (10)  and  (4)  are  identical  and  a  single  solution  is 
impossible.  Equations  (i),  (2)  and  (3)  represent  but  two  inde- 
pendent equations  and  are  insufficient  for  the  solution  of  three 
unknown  quantities. 


SEC.  IV  EQUILIBRIUM    OF    COPLANAR   FORCES  17 

(b)  Graphic  Method.  —  The  magnitudes  of  A,  B  and  C  are 
determined  by  closing  the  magnitude-direction  diagram. 
Choosing  the  convenient  scale-ratio  of  40  Ib.  to  i  in.,  we  lay  off 
oa  and  ab  (Fig.  4^)  representing  respectively  the  forces  E  and  D 
in  magnitude,  direction  and  sense.  The  diagram  is  to  be  closed 
from  b  to  o  by  three  lines  which  are  drawn  parallel  respectively 
to  the  directions  of  A  ,  B  and  C. 

Through  the  point  b  draw  a  line  parallel  to  the  direction  of 
either  A  ,  B  or  C  (let  us  say  C)  ;  and  through  the  point  o  draw  a 
line  parallel  to  the  direction  of  one  of  the  two  remaining  forces 
(B,  for  example).  The  diagram  may  be  closed  by  drawing 
any  line  cd  parallel  to  the  direction  of  the  remaining  force  A  ; 
and  we  conclude,  as  in  the  algebraic  method,  that  a  single 
solution  is  impossible. 

13.  General  Case.  —  By  giving  opposite  signs  to  magnitudes 
of  opposite  sense,  the  H-  and  V-  equations  may  be  expressed 
as  follows: 


=  8<z+  126  +  64-60 
2F  =  C  +  8o+  150  -  56 


By  indicating  clockwise  moments  as  positive  and  counter- 
clockwise moments  as  negative,  the  algebraic  sum  of  the  mo- 
ments of  the  forces  about  any  point  F  is 


8a  +  126)  +  $qb  -  q(C  +  So  +  150)  -  6op 
=  p(Sa  +  126  +  64  -  60)  -  q(C  +  80  +  150  -  5&) 
=  (p  X  I#)  -  (q  X  ZPO 
If  I#  =  o 
and  IF  =  o 

then  (p  X  Z£T)  --  (q  X  IF)  =  o 

or  I.MF  =  o 

for  any  values  of  p  and  q.  Hence,  if  the  horizontal  magnitudes 
are  balanced  and  the  vertical  magnitudes  are  balanced,  the 
moments  of  the  magnitudes  are  also  balanced  about  any  point 
chosen  as  the  center  of  moments;  and  any  equation  of  the  M- 
type  will  be  dependent  upon  the  H-  and  F-equations. 


1 8  THEORY   OF   FRAMED    STRUCTURES  CHAP.  I 

Suppose  that  ZF  =  o                                                   (5) 

and  ZMF  =  (p  X  Z#)  -  (q  X  IF)  =  o      (6) 

it  follows  that  p  X  2H  =  o 

If  #  =  o 

then  Eqs.  (5)  and  (6)  are  identical,  and  Z#  may  or  may  not 
equal  zero.  But  if  p  does  not  equal  zero,  as  when  the  center  of 
moments  is  not  in  line  with  the  horizontal  forces,  then 

•LH  =  o  (7) 

and  Eq.  (7)  is  dependent  upon  (5)  and  (6). 

Similarly  Eq.  (5)  is  dependent  upon  (6)  and  (7)  when  the 
center  of  moments  is  not  in  line  with  the  vertical  forces. 

Again  if  ZM  =  o  about  two  points  F  and  G  (Fig.  46)  then 

ZMF  =  o 

and  ZAf0  =  o 

or  pXZH  =  qXZV  (8) 

and  s  X  Z#  =  t  X  IF  (9) 

If  the  two  points  F  and  G  are  in  a  straight  line  with  the  point  of 
concurrency,  then 

p  =  q 
s       t 

and  Eqs.  (8)  and  (9)  are  identical  and  not  independent.  If 
the  two  points  F  and  G  are  not  in  a  straight  line  with  the  point 
of  concurrency,  then 

-  does  not  equal  - 
s  t 

or  pt  —  qs  does  not  equal  zero 

in  which  case,  Eqs.  (8)  and  (9)  are  independent. 
Multiply  Eq.  (8)  by  t,  and  Eq.  (9)  by  q 
pt  X  Z#  =  qtIV 
qsXXH  =  g/ZF 

(pt  -  qs)ZH  =  o  (10) 

Since  pt  —  qs  does  not  equal  zero,  then  Z.H  in  Eq.  (10)  must 
equal  zero.  But  if 

XH  =  o 
then  from  (8)  or  (9) 

ZF  =  o 


SEC.  IV  EQUILIBRIUM    OF    COP  LAN  AR   FORCES  IQ 

Hence  two  equations  of  the  type  ZAf  =  o  are  independent  if  the 
two  points  representing  the  centers  of  moments  are  not  in  the 
same  straight  line  as  the  point  of  concurrency;  and  any  addi- 
tional equation  of  either  the  H-,  V-  or  If -type  will  be  dependent 
upon  them. 

14.  Only  two  independent  statements  or  equations  can  be 
written  for  any  concurrent  system  of  coplanar  forces  in  equi- 
librium, and  it  follows  that  if  a  single  solution  is  possible  the 
unknown  elements  in  the  system  cannot  be  more  than  two. 

15.  Illustrative  Problem. — Assuming  that  the  known  ele- 
ments in  Fig.  40  are  the  same  as  before,  let  us  suppose  that  the 
magnitude  of  C  is  10  lb.,  and  that  the  sense  is  upward. 

(a)  Algebraic  Method. — Equate  the  horizontal  components 
of  opposite  sense. 

60  =  64  +  80  +  126 

Equate  the  vertical  components  of  opposite  sense. 

10  +  80  +  150  =  56 
Solve  for  a  and  b. 

a  =  -  5  lb. 
b  =  3  lb. 

Therefore  Ah  =    8#  =  —  40  lb., 

Av  =  150  =  -75  lb., 
and  A    =  170  =  —85  lb. 

The  negative  signs  indicate  that  our  assumption  regarding  the 
sense  of  the  force  A  and  its  components  was  wrong;  i.e.,  the 
force  acts  toward  the  point  0  instead  of  away  from  it. 
Likewise  Bh  =  126  =  36  lb., 

Bv  =    56  =  15  lb., 

and  B    =  136  =  39  lb. 

The  signs  are  positive  and  our  assumption  as  to  the  sense  of  B 
was  correct.  The  numerical  values  of  the  components  should 
now  be  indicated  in  a  sketch  similar  to  Fig.  46,  and  an  arith- 
tic  check  made  to  certify  that  the  components  balance. 

(b)  Graphic  Method. — Returning  to  the  point  b  (Fig.  4^)  we 
lay  off  be'  =   10  lb.  to  represent  the  force  C;  draw  c'd'  parallel 
to  the  direction  of  A,  and  d'o  parallel  to  the  direction  of  B. 
Or,  we  may  draw  c'e  parallel  to  the  direction  of  B  and  eo  parallel 


20  THEORY   OF   FRAMED   STRUCTURES  CHAP.  I 

to  the  direction  of  A .  The  magnitude  and  sense  of  each  force 
are  thus  definitely  determined;  in  the  first  process  by  the  lines 
c'd'  and  d'o;  in  the  second,  by  the  lines  c'e  and  eo.  By  scaling 
either  set  of  lines  we  find  that  the  magnitudes  of  A  and  B 
respectively  are  85  lb.,  acting  toward  the  point  0;  and  39  lb., 
acting  away  from  the  point  0. 

16.  Illustrative  Problem. — Now  let  us  suppose  that  all  'the 
elements  of  the  forces  in  the  system  which  we  have  been  dis- 
cussing are  known  except  the  directions  (including  sense)  of 
the  forces  A  and  C,  which  are  to  be  determined. 

(a)  Algebraic  Method. — We  shall  assume  that  the  direction 
and  sense  of  each  force  A  and  C  (Fig.  4^)  are  toward  the  right 
and  upward;  and  let 

c  represent  the  horizontal  component  of  C, 
d  represent  the  vertical  component  of  C, 
e  represent  the  horizontal  component  of  A , 

and          /  represent  the  vertical  component  of  A . 

Then  60  =  64  +  36  +  c  +  e 

and  80  +  d  +f  =  15 

The  magnitudes  of  A  and  C  are  85  lb.  and  10  lb.  respectively, 
and  their  components  are  rectangular;  hence 

c2  +  d2  =  io2 
and 

e*+f2  =  852 

Solving  the  four  equations  for  c,  d,  e  and  /,  we  find 

Ch  =  c  =  o  or        8.927 

Cv  =  d  =  io  or        4.507 

Ah  =  e  =  —40  or  —48.927 

and                         Av  =  f  =  —75  or  —69.507 

Each  component  has  two  values.  The  first  set  of  values  was 
anticipated  from  our  previous  experience  with  the  problem ;  but 
an  arithmetic  check  will  certify  that  the  second  set  is  equally 
true.  The  direction  and  sense  of  each  force  are  given  by  the 
ratio  and  signs  of  its  components  in  each  set. 

(b)  Graphic  Method. — Lay  off  in  Fig.  40  the  magnitudes  and 
directions  of  B,  E  and  D,  from  a  to  d.     About  a  and  d  as  centers, 


SEC.  IV 


EQUILIBRIUM    OF    COPLANAR   FORCES 


21 


draw  circumferences  having  radii  equal  respectively  to  the 
magnitudes  of  A  and  C,  intersecting  at  e  and  /.  The  slopes 
of  the  lines  de  and  df  represent  the  directions  of  the  force  C; 
similarly,  the  lines  ea  and  fa  represent  the  directions  of  the 
force  A . 

Whenever  an  unknown  element  is  the  direction  of  a  force 
having  a  known  magnitude,  the  algebraic  method  involves  the 
solution  of  a  quadratic  equation  which  may  have  two  real 
roots — as  in  the  present  case — one  real  root  or  two  imaginary 
roots.  Graphically,  the  known  magnitude  will  be  represented 
by  the  radius  of  a  circle,  its  circumference  having  two  points, 
one  point  or  no  point  in  common  with  either  another  circum- 
ference or  a  straight  line. 

17-  Special  Case. — The  system  of  forces  represented  in  Fig. 
5  represents  a  special  case.  The  magnitudes  of  three  forces, 


FIG.  5. 

A,  C  and  D  are  unknown.  Since  A  and  C  have  the  same 
direction,  the  magnitude  of  D  for  equilibrium  may  be  deter- 
mined algebraically  by  equating  the  components  normal  to  the 
direction  of  A  and  C.  The  normal  component  of  the  force 
B  is 

gk  =  -  X  240  =  192 
o 

The  force  D  has  a  slope  of  3  vertical  to  8  horizontal,  and  may  be 
resolved  into  the  components  ab  and  be  parallel  and  normal  to 


THEORY  OF  FRAMED  STRUCTURES        CHAP.  I 

the  line  AC  by  the  triangle  abc.     The  triangles  abd,  bee  and  jkg 
are  similar. 

Let  ad  =  3 

then  db  =  4 

and  ab  =  5 

Let  6e  =  30; 

then  ec  =  4^ 

and  &c  =  5# 

But  4*  -  3:3*  +  4:13:8 

therefore  be  =  $x  =  7.83 


and  ac  =  Vs2  +  y.832  =  9.29 

equating  the  normal  components, 

be  =  gk  =  192 

Since  ac  :  be  :  :  9.29  :  7.83 

the  magnitude  of  the  force  D  is 

9.29 
ac  =  —  X  192  =  227.8 

In  the  graphic  solution,  we  assume  any  magnitude  and  either 
sense  for  the  force  A  as  represented  by  pq,  and  lay  off  qs  =  240  : 
draw  st  parallel  to  the  direction  of  C,  and  tp  parallel  to  the  direc- 
tion of  D.  Then  any  line  tp  represents  the  magnitude  and  sense 
of  the  force  D.  The  magnitudes  of  A  and  C  cannot  be  deter- 
mined, but  their  difference  is  represented  by  the  difference  in  the 
lengths  of  pq  and  ts. 

Hence,  if  a  concurrent  system  of  forces  in  equilibrium  has 
three  unknown  magnitudes,  two  of  which  have  the  same  direc- 
tion, the  magnitude  of  the  third  may  be  determined;  but  the  two 
magnitudes  having  the  same  direction  cannot  be  determined. 

18.  Conclusion.  —  We  may  conclude  that  in  a  concurrent 
system  of  coplanar  forces  in  equilibrium,  the  unknown  elements 
cannot  be  determined  if  they  are  more  than  two  in  number. 
The  element  of  location  is  known  for  each  force;  hence  each 
unknown  element  will  be  either  a  magnitude  or  a  direction,  and 
will  occur  in  one  of  the  four  following  combinations: 

1.  Magnitude  and  direction  of  one  force. 

2.  Magnitudes  of  two  forces. 

3.  Directions  of  two  forces. 


SEC.  IV 


EQUILIBRIUM   OF   COPLANAR  FORCES 


4.  Magnitude  of  one  force  and  direction  of  another  force. 

Case  (2)  cannot  be  solved  if  the  two  unknown  magnitudes 
have  the  same  direction.  Cases  (3)  and  (4)  may  have  two 
solutions,  one  solution,  or  no  solution. 

19.  Problems. — Solve  the  following  problems  by  graphic  and 
algebraic  methods. 


'2000' 


1.  In  Fig.  6  four  concurrent  forces  are  in  equilibrium.     Find  the  magnitude  and 
direction  of  the  fourth  force. 

2.  Find  the  magnitudes  of  the  forces  P  and  Q  (Fig.  7)  for  equilibrium. 

3.  Five  concurrent  forces  are  in  equilibrium  (Fig.  8).     The  magnitudes  of  two 
of  the  forces  P  and  Q  are  520  Ib.  and  340  Ib.  respectively.     What  are  the  direc- 
tions of  P  and  Q? 

4.  Four  concurrent  forces  are  in  equilibrium  (Fig.  9).     The  magnitude  of 
Q  is  300  Ib.     Find  the  magnitude  of  P  and  the  direction  of  Q. 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  I 


5.  In  Fig.  10  the  moments  of  all  the  forces  are  balanced  about  the  two  points 
A  and  B,  i.e., 

ZMA  =  o 

and  *LMs  —  o 

yet  the  system  is  not  in  equilibrium  because  neither  the  horizontal  nor  the  vertical 
magnitudes  are  balanced.     Explain. 

6.  In  Fig.  n,  AB  represents  the  mast  of  a  derrick  40  ft.  long.     The  boom 
AC  is  60  ft.  long.     The  length  of  the  cable  BC  can  be  varied  to  place  the  boom 
in  any  desired  angle.     The  load  at  C  is  2,000  Ib.     Three  forces  at  C  are  in  equi- 
librium.    What  is  the  force  acting  along  the  direction  AC? 

SEC.  V.     COPLANAR  PARALLEL  FORCES 

20.  General  Considerations. — In  a  concurrent  system  the 
location  of  each  force  is  known  and  we  are  interested  in  the 
elements  of  magnitude  and  direction  only;  in  a  parallel  system 


CK--  a  --- 


FIG.  i2&. 


the  direction  of  each  force  is  known  and  we  have  the  elements 
of  magnitude  and  location  for  our  consideration. 

If  a  body  is  subject  to  the  action  of  a  system  of  parallel 
forces  which  have  vertical  directions,  the  condition 


=  o 

is  satisfied  whether  the  body  is  in  equilibrium  or  not.  Conse- 
quently this  equation  lends  no  aid  in  finding  the  unknown 
elements  for  the  equilibrium  of  such  a  system.  We  have  the 
equation 

IF  =  o 
and  equations  of  the  type 

ZAf  =  o 

for  the  algebraic  solution  of  a  system  of  vertical  forces. 

Let  us  examine  the  expressions  for  ZF  and  IMC  in  connec- 
tion with  Figs.  120  and  126  and  note  the  difference. 


SEC.  V  EQUILIBRIUM    OF    COPLANAR   FORCES  25 

The  algebraic  sum  of  the  magnitudes  of  the  three  forces  in 
Fig.  i2a  (indicating  magnitudes  of  upward  sense  as  positive)  is 

IF  =  +5   —  12  —  2  =  —  gib., 

i.e.,  the  resultant  of  the  three  forces  has  a  magnitude  of  9  Ib. 
acting  downward.  The  algebraic  sum  of  the  moments  of  the 
three  forces  about  the  point  C  (indicating  clockwise  moments 
as  positive)  is 

ZMC  =     —  50  +  1 2  (a  +  6)  +  2  (a  +  9) 
=   (  —  5  +  12  +  2)  a  +  90 
=     -a(IF)    +90 

Since  SV  does  not  equal  zero,  the  numerical  value  of  ZAfc, 
varying  with  the  distance  a,  is  different  for  any  two  points  not 
in  a  line  parallel  with  the  three  forces.  For  example,  when  C  is 
6  ft.  to  the  left  of  the  5-lb.  force 

a  =  +6 

and  ZMC  =  —  6(  — 9)  +  90  =  144  ft.-lb. 

When  C  is  3  ft.  to  the  right  of  the  2-lb.  force 

a  =  —12 
and  Z.MC  =  i2(  —  9)  +  90  =  — 18  ft.-lb. 

The  algebraic  sum  of  the  magnitudes  of  the  three  forces  (Fig. 
126)  is 

IF  =  4-  12 +  8  =  0 

The  algebraic  sum  of  the  moments  of  the  three  forces  about  the 
point  C  is 

ZMC  =  -40  +  i2(a  +  2)  -  8(fl  +  6) 
=  —(+4  —  12  +  8)0  —  24 
=  -<z(I  F)  -  24 
but  ZF  =  o 

therefore       ZMC  =  —  24 

has  the  same  numerical  value  for  any  distance  a. 
Hence,  in  a  system  of  parallel  forces,  if 

ZM  =  o 

about  some  point  chosen  as  the  center  of  moments,  it  does  not 
follow  that 

=  o 


26 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  I 


about  every  other  point,  as  illustrated  by  Fig.  120. 

But  when  IF  =  o 

as  in  Fig.  126,  then  SAT  has  the  same  numerical  value   for  all 

points  chosen  for  the  center  of  moments.     Hence  in  any  case 

where 

IF  =  o 
if  IM  =  o 

for  any  one  point,  it  follows  that 

IM  =  o 
for  all  points  and  equilibrium  is  assured.     There  can  be  but  one 


«-- 


independent  moment  equation  when 

IF  =  o 

Let  us  suppose  that  the  magnitudes  and  locations  of  the  five 
vertical  forces,  acting  upon  the  body  illustrated  in  Fig.  13,  are 
such  that 

ZMP  =  o 
and  IMG  =  o 


when  F  and  G  are  any  two  points  not  in  a  line  parallel  to  the 
system.  The  two  equations  signify  that  the  moments  of  the 
five  forces  are  balanced  about  F  and  G,  or 

(k  +  a)  A  +  (k  +  b)B  +  (k  +  c)C  =  (k  +  d)D  +  (k  +  e)E  (i) 
and  aA  +  bB  +  cC  =  dD  +  eE  (2) 


Subtract 

whence 

or 


kA  +  kB  +  kC  =  kD  +  kE 
A+B  +  C  =  D  +  E 

Z-F  =  o 


(3) 


SEC.  V 


EQUILIBRIUM   OF   COPLANAR   FORCES 


Hence,  if  the  moments  of  the  forces  in  a  parallel  system  balance 
about  any  two  points  not  in  a  line  parallel  to  the  system,  the 
magnitudes  will  also  balance  and  equilibrium  is  assured. 


\20*  \Q  fyf* 

<-5'--^< ^^^-vK^- 


Only  two  of  the  Eqs.  (i),  (2)  and  (3)  are  independent,  since 
each  may  be  derived  from  the  other  two. 

21.  Illustrative  Problems. — (i)  Seven  vertical  forces  hold 
the  body  (Fig.  140)  in  equilibrium.  The  magnitudes  and 


28  THEORY   OF   FRAMED   STRUCTURES  CHAP.  I 

locations  of  six  forces  are  known:  the  magnitude  and  location 
of  the  seventh  force  P  is  desired. 


T 

20  15 

16  25 

24  12 

60  52 

jf! 

8 

ZM  about  the  line  AB 

o  o 

15X0=      o  20X5  =  I0° 

25  X  8    =  200  16  X  14  =  224 

12  X  31  =  372  24  X  19  =  456 

572  780 

572 

8)208 

26 

The  sum  of  the  known  magnitudes  acting  upward  is  60  lb.,  and 
the  sum  of  the  known  magnitudes  acting  downward  is  52  lb.  In 
order  to  balance  the  magnitudes,  P  must  act  downward  having 
a  magnitude  of  8  lb.  Since  any  point  may  be  chosen  as  the 
center  of  moments,  it  is  expedient  to  select  a  point  in  the  line 
of  one  of  the  forces,  thereby  eliminating  that  force  from  the 
moment  calculations.  The  sum  of  the  counter-clockwise 
moments  about  the  line  AB  is  208  ft.-lb.  greater  than  the  sum 
of  the  clockwise  moments;  consequently  the  body  will  rotate 
counter-clockwise  unless  the  magnitude  8  lb.  acting  downward 
is  located  26  ft.  to  the  right  of  the  line  AB  in  order  to  balance 
the  moments. 

2.  The  location  of  Q  (Fig.  14^)  may  be  found  by  balancing 
the  moments  about  P.  The  magnitude  of  P  may  be  determined 
by  balancing  the  magnitudes;  or  by  balancing  the  moments 
about  any  point  not  in  the  line  of  Pt  after  the  location  of  Q  has 
been  found. 


SEC.  V        EQUILIBRIUM  OF  COPLANAR  FORCES  2Q 

3.  In  Fig.  i4c  two  magnitudes  are  unknown.     The  magnitude 
of  Q  may  be  found  by  balancing  the  moments  about  any  point 
in  the  line  of  the  force  P.     The  magnitude  of  P  may  be  found  by 
balancing  the  moments  about  any  point  in  the  line  of  the  force 
Q,  or  by  balancing  the  magnitudes  after  the  magnitude  of  Q  has 
been  determined. 

4.  In  Fig.  i4d  the  locations  of  two  forces  P  and  Q  are  un- 
known.   Let  a  and  b  represent  the  distances  from  the  line  A 
of  the  forces  P  and  Q  respectively.     The  magnitudes  balance, 
or 

IF  =  o 

regardless  of  the  values  of  a  and  b.  Let  us  attempt  a  solution 
by  writing  two  moment  equations  involving  a  and  b  —  the  one 
about  the  line  A  and  the  other  about  the  line  B. 


25  X    a  =  2$a  20  X    b  =  2ob 

8  X  26  =  208  16  X  14  =  224 

12  X  31  =  372  24  X  19  =  456 

580  +  2$a  680  +  206 

or  250  —  2ob  =  100 


24  X  12  =      288  8  X  5  =      40 

16  X  17  =      272  25(31  -  a)  =    775  -  250 

20(31  —  b)  =     620  —  2ob      15  X  31        =    465 

1,180  —  2ob  =  1,280  —  250 

or  250  —  206  =  100 

The  two  moment  equations  are  identical,  and  a  single  solution 
for  the  locations  of  the  forces  P  and  Q  is  impossible.  This 
conclusion  might  have  been  anticipated.  If  each  of  the  two 
unknown  elements  is  a  location,  then  all  magnitudes  are  known 
and  must  balance  for  equilibrium.  Since  there  can  be  but  one 
independent  moment  equation  when 

IF  =  o; 

it  follows  that  the  unknown  elements  cannot  be  determined  if 
there  is  more  than  one  unknown  location. 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  I 


22.  Conclusion. — We  may  conclude  that  in  a  parallel  system 
of  coplanar  forces  acting  upon  a  rigid  body  in  equilibrium,  the 
unknown  elements  cannot  be  determined  if  they  are  more  than 
two  in  number  or  represent  more  than  one  location.     The  ele- 
ment of  direction  is  known  for  each  force;  hence  each  unknown 
element  will  be  either  a  magnitude  or  a  location,  and  will  occur 
in  one  of  the  three  following  combinations: 

1.  Magnitude  and  location  of  one  force. 

2.  Magnitude  of  one  force  and  location  of  another  force. 

3.  Magnitude  of  two  forces. 

For  concurrent  systems  the  graphic  solution  is  generally 
shorter  than  the  algebraic.  There  is  very  little  to  recommend 
the  use  of  the  graphic  method  in  the  solution  of  a  parallel  sys- 
tem. The  algebraic  method  invariably  gives  the  shorter  solu- 
tion. We  shall  refer  to  the  graphic  method  in  connection  with 
parallel  forces  in  Section  VI. 

SEC.  VI.     COPLANAR  NON-CONCURRENT  NON-PARALLEL  FORCES 

23.  Transformed  System. — A  coplanar  non-concurrent  non- 
parallel  system  may  contain  horizontal  forces,  vertical  forces 


3c 


FIG.  15. 

and  inclined  forces;  and  any  unknown  element  in  the  system 
may  be  a  magnitude,  a  direction  or  a  location.  Such  a  system 
may  be  so  transformed  as  to  contain  only  horizontal  and  vertical 
forces,  by  resolving  each  inclined  force  into  its  horizontal  and 
vertical  components.  This  transformation  will  change  the 
character  but  not  the  number  of  the  unknown  elements,  and  is 
illustrated  in  Fig.  15. 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR   FORCES  31 

The  force  located  at  A ,  having  an  unknown  magnitude  and 
direction,  may  be  replaced  by  the  two  unknown  magnitudes 
a  and  b  having  known  directions  and  assumed  senses.  When 
the  magnitudes  a  and  b  with  proper  senses  have  been  determined, 
the  magnitude,  direction  and  sense  of  the  original  force  may  be 
found. 

The  force  at  B,  having  an  unknown  magnitude,  may  be 
replaced  by  the  two  components  3^  and  5^  having  only  one 
unknown  quantity  in  the  expressions  which  represent  their 
magnitudes. 

The  magnitude  of  the  force  at  C  is  75  lb.,  but  its  direction  is 
unknown.  Let  d  and  e  represent  the  magnitudes  of  the  two 
components  with  senses  assumed.  In  this  instance  one  un- 
known direction  has  been  replaced  by  two  unknown  magnitudes ; 
but  one  of  the  magnitudes  may  be  expressed  in  terms  of  the 
other,  since 

d2  +  e2  =  752 

The  magnitude  and  direction  of  the  force  D  are  known,  but 
the  location  is  unknown.  It  is  desired  to  locate  the  force  with 
reference  to  the  point  E.  Resolve  the  force  into  its  horizontal 
and  vertical  components.  Their  magnitudes  are  80  lb.  and  60 
lb.  respectively.  Locate  arbitrarily  one  of  the  components. 
For  example,  assume  that  the  horizontal  component  acts 
through  the  point  E.  Let/  represent  the  distance  of  the  verti- 
cal component  from  E.  The  intersection  of  the  two  compo- 
nents at  0  marks  a  location  of  the  force  D.  The  point  O  will 
be  found  on  the  left  or  the  right  of  the  point  E}  as  the  solution 
gives  a  positive  or  a  negative  numerical  value  for  /.  Many 
different  points  0  may  be  located  by  assigning  different  loca- 
tions for  the  horizontal  component;  but  all  such  points  thus 
found  will  lie  in  the  same  straight  line,  and  serve  equally  well 
in  locating  the  force  D.  The  slope  of  this  line  will  represent  the 
direction  of  the  given  force. 

24.  Illustrative  Problem. — Let  Fig.  16  represent  a  trans- 
formed system  of  coplanar  non-  concurrent  non-parallel  forces, 
after  each  inclined  force  has  been  replaced  by  its  horizontal 
and  vertical  components.  We  wish  to  ascertain  how  many 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  I 


independent  equations  may  be  written  concerning  the  equi- 
librium of  the  rigid  body  upon  which  the  forces  are  acting.  For 
the  solution  of  the  four  unknown  magnitudes  A,  B,  C  and  D, 
we  shall  attempt  to  write  four  independent  equations,  i.e.  the 
H-  and  F-equations  and  two  M-equations,  one  about  the  point 
0  and  the  other  about  the  point  P. 

Z.H  =  A+B-2$  =  o  (i) 

IF  =  C  -  10  -  4  ~  D  =  o  (2) 

0  =  loC  +  12,4  +  40  -  400  +  loD  =  o  (3) 

=  —i6B  —  100  +  2oC  —  4^4  =  o  (4) 


Eliminate  B  and  C  from  (3)  and 
(4)  and  each  equation  reduces  to 

12  A   +  20D  =    220 

This  is  but  another  way  of  say- 
ing that  (4)  adds  nothing  which 
has  not  been  previously  stated 
by  (i),  (2)  and  (3). 

Write  anM-equation  for  some 
other  point  and  see  if  an 
identical  equation  cannot  be 

derived  from  Eqs.  (i),  (2)  and  (3). 
25.  General  Case.  —  Now  consider  the  general  case  illustrated 

in  Fig.  17. 


FIG.  16. 


«-,$---> 


<--&-*> 


71 > 


"STT 


FIG.  17. 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR    FORCES  33 


=  +F  +  G  -  E 

=  -(a  +  h)A  +  (b  +  ti)B  -  (c  +  k)C  +  (d  +  A)£ 

-(/•  +  £)F  -  (g  +  fc)G  +  (e  +  k)E 
-aA      +bB      -cC      +dD      -fF      -gG      +eE 
rP  -  ZM0  =  h(-A  +  B  -  C  +  Z>)  +  £(-F  -  G  +  -E) 

=  (A  X  ZF)  -  (&  X  ZF) 

If  ZF  =  o  (i) 

and  ZF  =  o  (2) 

then  ZAfp  —  ZJf0  =  o 

or  ZMP  =  ZM0 

for  any  value  oi  h  or  k. 

Hence,  if  the  vertical  magnitudes  are  balanced  and  the  hori- 
zontal magnitudes  are  balanced,  the  algebraic  sum  of  the 
moments  of  the  forces  is  the  same  for  all  points  chosen  as  the 
center  of  moments;  in  which  case  if 

J.M  =  o  (3) 

for  any  point,  the  moments  are  balanced  about  all  points, 
equilibrium  is  assured;  and  there  will  be  one  and  only  one 
independent  Af-equation  as  illustrated  in  Article  24. 

26.  Four  Groups. — The  three  statements  or  Eqs.  (i),  (2) 
and  (3)  are  the  necessary  and  sufficient  conditions  for  insuring 
the  equilibrium  of  a  body,  when  acted  upon  by  any  system  of 
coplanar  non-concurrent  forces;  and  any  fourth  equation  will  be 
dependent  on,  and  may  be  derived  from  them.  Situations  may 
arise  where  it  will  be  convenient  to  express  the  conditions  of 
equilibrium  in  other  forms.  The  four  ways  in  which  the  three 
equations  may  be  stated  are  as  follows: 

GROUP  i  GROUP  2  GROUP  3  GROUP  4 

ZF  =  o  1.Mp  —  o  ZA/p  =  o  J.Mp  =  o 

ZF  =  o  ZM0  =  o  ZM0  =  o  ZM0  =  o 

ZM  =  o  ZF  =  o  ZF  =  o  ZM0  =  o 

in  which  O,  P  and  Q  represent  different  points  in  the  plane  of  the 
forces.  The  fact  that  certain  restrictions  must  be  placed  upon 
the  relative  positions  of  these  points  in  the  last  three  groups,  in 


34 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  I 


Z.Mp  =  —12  +  63  —  6A  —  loB  —  $a  =  o 
Z.M0  =  63  —  48  —  ioB  —  50  +  24  =  o 


order  that  the  three  equations  may  be  independent,  is  illus- 
trated by  the  following  examples. 

Example  i . — Find  the  magnitudes  A  and  B  and  the  distance  a 
(Fig.  1 8)  for  equilibrium,  using  the  equations  of  group  2,  and 
choosing  any  two  points  P  and  0  in  a  vertical  line. 

•  « 

(2) 
(3) 

The  three  equations   are   not 
independent,    since   the   third 
fy  #  may  be  derived  by  subtract- 
ing the  second  from  the  first. 

Example  2. — Use  the  equa- 
tions of  group  2  and  see  if  a 
solution  is  possible  by  choos- 
ing any  other  two  points,  0 
andP: 


' 


^ — — — — 


-— -> 


4' 


FIG.   1 8. 


(a)  In  a  vertical  line. 

(b)  Not  in  a  vertical  line. 

Example  3. — Use  the  equations  of  group  3  and  see  if  a  solution 


10' 


f " 

5' 


--4- 

s^ 


8' 


FIG.  19. 

is  possible  by  choosing  any  two  points,  O  and  P : 

(a)  In  a  horizontal  line. 

(b)  Not  in  a  horizontal  line. 

Example  4. — Find  the  magnitudes  A  and  B  and  the  distance  a 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR   FORCES  35 

(Fig.  19)  for  equilibrium  using  the  equations  of  group  4  and 
choosing  three  points,  0,  P  and  Q  in  a  straight  line. 

=  80  —  100  +  5^4  —  80  +  aB  =  o  (i) 

0  =  20  —  (16  —  a)B  +  384  —  120  —  176  —  nA  =  o  (2) 
=  1  80  -  150  +  88  +  13^  -  130  +  (8  +  a)B 

-192  =  o  (3) 

Multiply  Eq.  (3)  by  2,  add  Eq.  (2),  and  divide  the  sum  by  3. 
The  quotient  is  Eq.  (i).  Only  two  independent  equations  are 
represented,  since  any  one.  of  the  three  may  be  derived  from  the 
other  two,  and  a  solution  of  the  problem  as  stated  is  impossible. 

Example  5.  —  Choose  three  points  O,  P  and  Q  not  in  a  straight 
line  and  see  if  a  solution  is  possible. 

It  has  been  shown  that  under  certain  conditions  the  three 
equations  in  each  of  the  last  three  groups  represent  but  two 
independent  equations,  and  are  insufficient  for  the  solution  of 
three  unknown  quantities.  The  general  cases  will  now  be 
considered. 

27.  Group  2.  —  Suppose  that  in  Fig.  17 

=  o  (i) 

=  o  (2) 

and  I#  =  o  (3) 

then       ZMp  -  ZM0  =  (h  X  IF)  -  (k  X  Z#)  =  o  (4) 

If  the  two  points  P  and  0  are  in  a  vertical  line 

h  =  o 
hence  h  X  IF  =  o 

for  any  value  of  Z  V,  in  which  case, 
(a)  from  Eq.  (4) 

k  X  I#  =  o 


and,  since  k  is  not  zero,  Z.H  =  o  and  it  is  obvious  that  Eq. 
(3)  may  be  derived  from  Eq.  (4)  which  was  derived  from  Eqs. 
(i)  and  (2)  and  is  dependent  upon  them.  Hence,  when  the  two 
points  P  and  0  are  in  a  vertical  line,  Eqs.  (i),  (2)  and  (3)  repre- 
sent but  two  independent  equations  (see  Examples  i  and  20). 

(b)  the  value  of  Z  V  may  or  may  not  equal  zero  and  equi- 
librium is  not  assured. 


36  THEORY   OF   FRAMED    STRUCTURES  CHAP.  I 

But  if  the  two  points  P  and  0  are  not  in  a  vertical  line,  h  in 
Eq.  (4)  does  not  equal  zero,  in  which  case 

to  k  X  ZH 

in   Eq.   (4)  may  or  may  not  equal  zero;  hence  I#  may  or 
may  not  equal  zero  and  Eq.  (3)  is  independent  of  Eqs.  (i)  and 

(2)  (see  Example  26). 
(d)  since  from  Eq.  (3) 

I#  =  o 
then  from  Eq.  (4) 

h  X  ZF  =o  and  since  h  does  not  equal  zero 
then  IF  =  o  (5) 

and  equilibrium  is  assured.     Equation  (5)  does  not  represent 
an  independent  equation  since  it  is  derived  from  (i),  (2)  and 

(3). 

28.  Group  3. — Similarly  it  may  be  demonstrated  that  Eq. 

(3)  is  dependent  upon  Eqs.  (i),  (2)  and  (5)  when  the  two  points 
P  and  0  are  not  in  a  horizontal  line  (see  Examples  30  and  36). 

29.  Hence  in  any  case  where  the  moments  of  all  the  forces  in 
a  non-concurrent  non-parallel  system  are  balanced  about  any 
two  points  in  a  diagonal  line,  if  the  horizontal  magnitudes  are 
balanced  the  vertical  magnitudes  are  also  balanced;  or  if  the 
vertical  magnitudes  are  balanced,  the  horizontal  magnitudes 
are  also  balanced.     Only  the  H-  or  the    F-equation  can  be 
independent  of  the  two  moment  equations. 

30.  Group  4. — Suppose  that  in  Fig.  17 

Z.MP  =  o  (i) 

ZM0  =  o  (2) 

and  2MQ  =  o  (6) 

then        ZMP  -  IM0  =  (h  X  IF)  -  (k  X  Z#)  =  o  (7) 

and         ZM0  -  IM0  =  (s  X  IF)  -  (p  X  I#)  =  o  (8) 

If  the  three  points  O,  P  and  Q  are  in  a  straight  line 

h  _  k 
s  ==  p 

and  Eqs.  (7)  and  (8)  are  identical  and  not  independent  (see 
Example  4).     Eliminate  IF  in  (7)  and  (8) 
k  X  Iff       p  X 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR   FORCES  37 

or  h  X  2.H   _  k  X  2.H  ,  , 

If  the  three  points  O,  P  and  ()  are  not  in  a  straight  line,  then 

h  ,  ,k 

-  does  not  equal  — 
s  p 

in  which  case  Eqs.  (7)  and  (8)  are  independent;  and  1H  in 
Eq.  (9) must  equal  zero  (see  Example  5).     But  if 


then  from  Eq.  (7)  or  (8) 


=  o 


=  o 


31.  Hence  in  any  case  where  the  moments  of  all  the  forces 
in   a  non-concurrent  non-parallel  system   are  balanced  about 
any  three  points  not  in  a  straight  line,  the  horizontal  magni- 
tudes are  balanced,  the  vertical  magnitudes  are  balanced,  and 
the  H-  and   F-equations  will  be  dependent  upon  the  three 
If  -equations. 

32.  One  Unknown  Location.  —  In  Fig.  20  the  unknown  loca- 


b 
A-*- 


->K 

fc* 


-> 


4' 

t 
s' 


FIG.  20. 

tions  of  two  forces  are  represented  by  the  quantities  b  and  c. 
Let  the  horizontal  magnitude  A  represent  a  third  unknown 
element.  All  the  vertical  magnitudes  are  known  and  balanced. 
The  magnitude  A,  determined  by  balancing  the  horizontal 
magnitudes,  equals  6  lb.,  and  acts  to  the  right.  Since  the 
magnitudes  are  now  balanced  horizontally  and  vertically,  it  is 
possible  to  have  only  one  independent  Af-equation  for  the 
solution  of  b  and  c,  and  a  single  solution  is  impossible. 


38  THEORY    OF   FRAMED    STRUCTURES  CHAP.  I 

33.  Show  that  a  similar  condition  is  encountered  when  the 
unknown  elements  are  two  locations  and  one  direction.     There 
can   be   only   one   unknown  location  if  a  single  solution  is 
possible. 

34.  Combinations. — The  unknown  elements  which  can  be 
wholly  or  partially  determined  in  a  coplanar  non-concurrent 
non-parallel  system  of  forces  will  appear  in  one  of  the  following 
combinations;  where  P,  Q  and  S  represent  any  forces  which 
have  unknown  elements 


1.  The  magnitude  of  P,  the  magnitude  of  Q  and  the  magnitude  of  S. 

2.  The  direction  of  P,  the  direction  of  Q  and  the  direction  of  S. 

3.  The  magnitude  of  P,  the  magnitude  of  Q  and  the  direction  of  S. 

4.  The  magnitude  of  P,  the  magnitude  of  Q  and  the  direction  of  Q. 

5.  The  magnitude  of  P,  the  magnitude  of  Q  and  the  location  of  S. 

6.  The  magnitude  of  P,  the  magnitude  of  Q  and  the  location  of  Q. 

7.  The  direction  of  P,  the  direction  of  Q  and  the  magnitude  of  S. 

8.  The  direction  of  P,  the  direction  of.  Q  and  the  magnitude  of  Q. 

9.  The  direction  of  P,  the  direction  of  Q  and  the  location  of  S. 

10.  The  direction  of  P,  the  direction  of  Q  and  the  location  of  Q. 

11.  The  magnitude  of  P,  the  direction  of  Q,  and  the  location  of  5. 

12.  The  magnitude  of  P,  the  direction  of  P  and  the  location  of  P. 

13.  The  magnitude  of  P,  the  direction  of  P  and  the  location  of  Q. 

14.  The  magnitude  of  P,  the  direction  of  Q  and  the  location  of  P. 

15.  The  magnitude  of  P,  the  direction  of  Q  and  the  location  of  Q. 


Cases  (i),  (4)  and  (12)  are  more  frequently  encountered  in 
the  theory  of  framed  structures  than  any  or  all  of  the  others. 
Several  cases  are  indeterminate  under  certain  conditions. 
Case  (i)  is  in  this  class  when  P,  Q  and  S  have  the  same  location, 
or  are  parallel.  Other  cases  are  partially  indeterminate,  for 
they  may  have  several  solutions  or  none.  Any  problem  which 
involves  a  force  having  a  known  magnitude  and  an  unknown 
direction  is  in  this  class.  The  most  difficult  problems  are  to 
be  found  in  cases  (2)  and  (7). 

35.  Illustrative  Problems. — i  Determine  the  unknown  ele- 
ments necessary  for  equilibrium  (Fig.  21  a). 

(a)  Algebraic1  Method. — The  three  elements  of  each  of  the 

xThe  term  "algebraic"  is  used  simply  for  the  purpose  of  distinguishing  the 
method  of  computing  by  the  use  of  numerals  or  letters,  from  the  graphic  method 


SEC.  VI 


EQUILIBRIUM   OF    COPLANAR   FORCES 


39 


four   forces   shown   are   given.     The   elements   necessary   for 
equilibrium  are  the  magnitude,  direction  and  location  of  a  fifth 


\5^ 
Ch--Z40*> 


^          2f'  AS'\        10  f~ 

By*  120*  l7W-2*>* 


D^ioo# 


Cv*i$o 


H      'DV:24° 


Cvtuv 

FI6.2IC  *  * 


160 


FIG.2IG 


force.  The  problem  is  one  of  case  (12).  Make  a  sketch  (Fig. 
2ib)  and  transfer  to  it  the  horizontal  and  vertical  components 
of  the  four  known  forces  properly  located. 


in  which  quantities  are  represented  by  the  length,  direction  and  location  of  lines. 
Some  writers  refer  to  the  two  methods  as  the  analytic  and  graphic;  but  this  is 
manifestly  incorrect,  since  both  methods  are  analytic.  The  frequent  use  how- 
ever, of  the  word  "algebraic"  should  be  discouraged.  It  is  true  that  algebraic 
equations  have  been  used  in  the  discussions  which  have  preceded,  but  in  the 
actual  solution  of  problems  their  use  is  often  a  hindrance  rather  than  a  help. 
In  the  first  place,  if  the  equations  are  not  written,  the  problem  can  frequently 
be  solved  more  quickly  and  easily;  and  the  computations  can  be  arranged  in  a 
more  convenient  form  for  checking.  But  a  more  important  reason  for  discarding 
the  equation  whenever  possible  is  the  fact  that  by  so  doing,  the  attention  is 
held  fast  to  the  original  problem  in  statics,  and  not  drawn  far  afield  by  an 
exercise  in  elementary  algebra. 


4O  THEORY    OF   FRAMED    STRUCTURES  CHAP.  I 

The  horizontal  and  vertical  components 


of  A  are  160  Ib.  and  300  Ib.  respectively, 
15  =  I7  an(*  are  S°  Placec*  on  tne  sketch  as  to 


=   64 

=  I7 

intersect  at  a  point  in   the  line  of  the 

x   8  =  l6°       force  ^4.     The  components  of  the  other 
known  forces  are  determined  and  located 


340 

Av  =  77  x  I5  =  3°°      in  a  similar  manner.     (The  sketch  in  Fig. 
2ic  is  superfluous,  except  that  it  shows  one 
of  the  several  other  forms  in  which  the   components   might 
have  been  correctly  placed.) 

The  sum  of  the  horizontal  components  acting  to  the 

160    240       right  is  greater  by  180  than  the  sum  of  the  horizontal 

1  60  components    acting    to    the  left;  consequently  the 

body   will   move   to   the    right  unless  a  horizontal 

component  having  a  magnitude  of  180  acts  to  the 

left  to  balance  the  horizontal  forces. 
1  80 

Z.H  A    vertical    component    having    a    magnitude    of 

i  240    must    act    downward    to    balance    the    vertical 

180    3<X  f°rces-    Locate  one  of  the  components  arbitrarily. 

24o  Let  us  say  that  the  vertical  component  acts  at  the 

540    300  upper  right-hand  corner  of  the  body  as  shown  in 

3°°  Fig.  2id.     Find  the  algebraic  sum  of   the  moments 

240  of  the  magnitudes  about  any  point. 

The  point  0  is  chosen,  for 

O         ZM0  O  the  moments  of  the  three 

5  X  ioo  =       800     25  X  180  =    4,500 

10  x  160  =    1,600    30  x  240  =    7,200      magnitudes      which      pass 

2  x  300  =      600  through     it     are     thereby 

4o  x  240  =  _Q,6oo  _       eliminated.     The  clockwise 

II'7°°       moments   about  the  point 

jijpb  °  are  greater  by  900  ft.- 

5  Ib.  than  the  counter  clock- 

wisemoments.      The   body 

will  rotate  clockwise  about  the  point  O  unless  the  horizontal 
component  with  its  magnitude  of  180,  which  must  act  to  the 
left,  passes  through  the  point  E  at  a  distance  5  above  the  point 
O  (Fig.  2ie).  Check  the  computations.  See  that  the  magni- 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR   FORCES  41 

tudes  are  balanced  horizontally  and  vertically,  and  that  the 
moments  are  balanced  about  some  point  other  than  O. 

The   magnitude  of  the  required  force 

8°2  I  32'^°°  is  300;  its  sense  is  downward  to  the  left; 

\/~~oaoo  =    oo        ^  *s  l°cated  by  the  point  E;  and  its  direc- 
tion FEj  determined  by  the  ratio  of  its 
components,  is  4  vertical  to  3  horizontal.     The  Figs.  2ic,  d  and 
e   are   unnecessary   for   the    solution;  but  were    sketched  to 
illustrate  the  successive  steps  in  the  solution. 

Instead  of  locating  the  vertical  component  at  the  upper  right- 
hand  corner,  locate  the  horizontal  component  at  some  point  and 
determine  the  location  of  the  vertical  component.  Note  that 
the  two  components  thus  located  intersect  at  some  point  in  the 
line  EF. 

Graphic  Method. — Since  a  concurrent  system  of  forces  cannot 
result  in  rotation  about  the  point  of  concurrency,  equilibrium 
is  assured  if  the  magnitude-direction  diagram  is  a  closed  polygon 
and  the  sense  is  continuous.  While  the  closed  magnitude-direc- 
tion diagram  is  a  necessary  condition  for  equilibrium  in  a  non- 
current  system,  it  is  not  a  sufficient  condition.  The  closed 
magnitude-direction  diagram  takes  no  account  of  the  location 
of  forces  and  insures  against  translation  only.  In  order  that 
there  may  be  no  rotation,  another  figure,  which  will  be  called 
the  location-direction1  diagram  must  also  be  drawn,  but  not 
necessarily  closed. 

1  One  of  the  two  diagrams  which  are  drawn  for  a  graphic  solution  of  a  system 
of  coplanar  non-concurrent  forces  is  usually  called  the  "force  polygon";  the 
other,  the  "equilibrium  polygon."  These  terms  do  not  seem  to  describe  ade- 
quately the  essential  qualities  of  the  two  diagrams.  Any  line  in  the  force 
polygon,  so  called,  does  not  completely  represent  a  force.  It  represents  only  two 
of  the  three  elements,  viz.,  magnitude  and  direction.  The  term  "equilibrium 
polygon  "  seems  to  be  applicable  equally  to  both  diagrams,  since  both  are  requisite 
for  equilibrium.  Any  line  in  the  equilibrium  polygon,  or  as  it  is  sometimes 
called,  the  funicular  polygon  or  string  polygon,  also  represents  but  two  of  the 
three  elements  of  a  force,  viz.,  location  and  direction.  Hereafter,  we  shall  use 
the  distinctive  terms,  magnitude-direction  diagram  and  location-direction 
diagram.  The  lines  in  the  first  diagram  will  be  called  magnitude-directions; 
in  the  second,  location-directions.  The  two  diagrams  have  nothing  in  common 
except  the  element  of  direction  and  the  sense,  which  are  registered  in  both. 
There  is  nothing  pertaining  to  locations  in  the  first,  and  nothing  pertaining  to 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  I 


The  closing  of  the  magnitude-direction  diagram  in  the  graphic 
method  corresponds  to  the  solution  of  I.H  =  o  and  IF  =  o 
in  the  algebraic  method;  while  the  drawing  of  a  location-direc- 
tion diagram  corresponds  to  the  solution  of  IM  =  o. 

The  body  in  Fig.  2ia  is  re-drawn  to  scale  (Fig.  2i/)  and  full 
lines  are  added  to  represent  known  location-directions.  The 


FlG.Zlg 


lengths  of  these  lines  have  no  significance  whatever,  since  they 
do  not  represent  magnitudes  in  any  particular.  The  magni- 
tude-direction diagram  is  now  constructed  (Fig.  2ig)  from  a  to 
e  by  laying  off  to  scale  the  known  magnitude-directions  in  any 
order  and  in  the  same  manner  as  for  a  concurrent  system.  The 
diagram  is  closed  by  drawing  the  magnitude-direction  ea  which 

magnitudes  in  the  second.  There  is  no  connection  between  the  scale  ratios 
used  in  laying  off  the  elements  in  the  two  diagrams;  for  in  the  first  the  scale  ratio 
is  pounds-to-the-inch,  in  the  second,  feet-to-the-inch. 


SEC.  VI  EQUILIBRIUM   OF   COPLANAR   FORCES  43 

gives  the  magnitude,  direction  and  sense  of  the  equilibrant  E. 
The  location  of  the  equilibrant  with  reference  to  the  body  (Fig. 
2 1/)  remains  to  be  determined.  This  may  be  accomplished  by 
drawing  the  location -direction  diagram  in  one  of  two  ways : 

(a)  By  combining  the  forces  in  pairs  and  drawing  the  location- 
directions  of  their  resultants. 

(b)  By  resolving  the  forces  and  drawing  the  location-direc- 
tions of  their  components. 

The  first  method  is  the  simpler  and  more  direct  way,  and  will 
now  be  considered. 

(a)  By  Drawing  the  Resultants. — The  magnitude-direction 
and  the  location-direction  of  a  force  are  designated  by  the  same 
letters,  placed  at  the  ends  of  the  magnitude-direction  and  on 
opposite  sides  of  the  location-direction.     Obviously  the  magni- 
tude-direction and  the  location-direction  of  any  force  are  alike 
in  sense  and  are  parallel.     The  magnitude-direction  of  the  re- 
sultant of  the  forces  A  and  B  is  ac ;  the  location-direction  of  this 
resultant  is  ac  drawn  through  /:  it  intersects  the  location- 
direction  of  the  force  D  at  g.     The  magnitude-direction  of  the 
resultant  of  the  forces  ac  and  D  is  ad;  the  location-direction  of 
this  resultant  is  ad  drawn  through  g:  it  intersects  the  location- 
direction  of  the  force  C  at  h.     The  magnitude-direction  of  the 
resultant  of  the  forces  ad  and  C  is  ae;  the  location-direction  of 
this  resultant  is  ae  drawn  through  h.     The  resultant  and  the 
equilibrant  of  a  system  of  forces  have  the  same  magnitude, 
direction  and  location;  they  differ  in  sense  only.     Hence,  the 
location-direction  of  the  equilibrant  E  is  the  line  ea,  and  any 
point  in  it  may  serve  as  a  location. 

Compare  the  results  of  the  graphic  and  algebraic  solutions. 

Draw  the  magnitude-direction  and  the  location-direction  ce. 
Explain  why  the  location-directions  ac  and  ce  intersect  on  the 
location-direction  ae. 

The  location-direction  diagram  is  not  a  closed  figure  when 
constructed  by  drawing  resultants. 

(b)  By  Drawing  the  Components. — The  method  of  drawing 
the  location-directions  of  the  resultants,  while  short  and  sim- 
ple, obviously  could  not  have  been  employed  if  the  directions  of 
the  forces  had  been  parallel;  for  there  would  have  been  no  points 


44 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  I 


of  intersection  in  the  location-direction  diagram.  The  method 
of  drawing  the  location-directions  of  the  components  of  forces 
has  a  wider  range  of  applicability  than  the  preceding  method, 
but  necessitates  the  drawing  of  more  lines.  The  procedure  for 
finding  the  magnitude,  direction  and  sense  of  the  equilibrant 


is  the  same  as  in  the  foregoing  solution;  only  in  the  method  of 
locating  the  equilibrant  are  the  two  solutions  different. 

The  full  lines  (Fig.  2iti)  represent  known  location-directions. 
The  magnitude-direction  diagram  has  been  closed  (Fig.  2ik). 
Choose  any  convenient  point  o,  called  the  pole;  and  draw  the 
magnitude-directions  oa,  ob,  oc,  od  and  oe.  The  pole  o  should 
not  be  taken  on  a  line  passing  through  any  two  adjacent  apexes. 
Select  another  convenient  point  on  a  known  location-direction, 


SEC.  VI  EQUILIBRIUM    OF    COP  LAN  AR   FORCES  45 

as  /  on  ab,  and  draw  the  location-directions  ob,  oc  and  od  as 
indicated;  thus  establishing  the  points  g,  h  and  j.  Through 
/  and  j  draw  the  location-directions  oa  and  oe  respectively. 
Their  intersection  at  k  closes  the  location-direction  diagram  and 
determines  the  location-direction  ea  of  the  equilibrant. 

The  three  forces  acting  at  /  are  in  equilibrium,  since  their 
location-directions  ab,  bo  and  oa  are  concurrent  and  their  magni- 
tude-directions ab,  bo  and  oa  form  a  closed  diagram.  The  con- 
current forces  acting  at  g,  h,j  and  k  are  also  in  equilibrium  for  a 
similar  reason.  The  two  equal  and  opposite  forces  acting  along 
the  line/g;  the  one  bo  toward/,  the  other  ob  toward  g,  are  bal- 
anced. The  same  is  true  with  respect  to  the  two  forces  acting 
along  each  of  the  other  sides  of  the  location-direction  diagram. 

Although  an  infinite  number  of  location-direction  diagrams 
may  be  drawn  by  choosing  different  points  for  the  pole  o  or 
the  starting  point/;  yet  in  each  diagram  thus  drawn,  the  loca- 
tion-directions ao  and  oe  will  invariably  intersect,  not  at  the 
same  point,  but  on  the  same  line.  This  line  represents  the 
locus  of  the  location-direction  ea. 

It  has  been  shown  that  when  Z.H  =  o  and  I  V  =  o,  if  1.M  = 
o  about  any  point,  thenZ  M  =  o  for  all  points.  Likewise  when 
the  magnitude-direction  diagram  closes,  it  follows  that  if  one 
location-direction  diagram  closes,  all  will  close. 

Some  care  must  be  exercised  in  selecting  the  points  o  and/, 
lest  the  location-direction  diagram  fall  beyond  the  limits  of  the 
drawing  paper. 

A  collapsible  frame  composed  of  five  members  and  having 
the  configuration  of  the  diagram  fghjk,  if  substituted  for  the 
rigid  body,  will  hold  the  five  forces  in  equilibrium.  The  equilib- 
rium will  be  unstable  however,  for  the  frame  will  collapse  if 
only  a  slight  change  be  made  in  its  shape.  The  same  is  true  of 
any  one  of  the  infinite  number  of  frames  which  may  be  con- 
structed by  choosing  different  points  o  and  /  in  this  or  in  any 
other  problem  in  which  the  frame  has  four  or  more  members, 
and  thereby  made  susceptible  to  a  collapse. 

Location-direction  Drawn  Through  Given  Points. — An  infi- 
nite number  of  location-direction  diagrams  may  be  drawn 
through  any  given  point  p  on  a  location-direction,  by  choosing 


46 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  I 


different  positions  for  the  pole  o  and  beginning  the  location- 
direction  diagram  at  p  in  each  instance.  An  infinite  number  of 
location-direction  diagrams  may  be  drawn  through  two  points, 
p  on  one  location-direction  and  s  on  another.  Call  the  loca- 
tion-directions on  which  the  two  points  p  and  5  are  located  xy 
and  yz,  and  draw  the  location-direction  oy  connecting  p  and  s. 
In  constructing  the  magnitude-direction  diagram,  arrange  the 
order  so  that  the  magnitude-directions  xy  and  yz  are  adjacent. 
Draw  the  magnitude-direction  oy  and  select  any  point  o  on  it 
for  the  pole. 

Three  points/,  g  and/  (Fig.  2im)  may  be  selected  at  random 


FIG^/n 


on  the  location-directions  of  three  forces;  and  a  location-direc- 
tion diagram  drawn  through  them,  provided  the  points  are  not 
in  a  straight  line.  Call  the  location-directions  on  which  /,  g 
and/  are  situated,  ab,  be  and  cd  respectively;  and  draw  the  loca- 
tion-directions ob  and  oc  through/,  g  and/  as  indicated.  Begin 
the  magnitude-direction  diagram  (Fig.  2 in)  so  that  be  will 
appear  between  ab  and  cd.  Draw  the  magnitude-directions 
ob  and  oc  intersecting  at  o,  the  desired  pole. 

Complete  the  magnitude-direction  diagram  in  two  ways; 
thus  giving  two  location-direction  diagrams  through  /,  g  and  /. 

Evidently  the  points/,  g  and/  may  be  connected  by  location- 
directions  in  two  other  ways;  viz.,  by  connecting/  and  g,  and 
/  and/;  or  by  connecting/  and/,  and  g  and/.  Thus,  several 


SEC.  VI 


EQUILIBRIUM    OF    COPLANAR   FORCES 


47 


constructions  of  the  location-direction  diagram  through  the 
three  given  points/,  g  and  j  are  possible;  but  the  number  for 
any  given  system  is  limited,  and  depends  upon  the  number  of 
forces  in  the  system. 

2.  Four  vertical  forces  ab,  bc7  cd  and  de  (Fig.  220)  are  com- 
pletely known.  The  direction  of  the  force  ef  is  vertical.  Find 
the  magnitude  of  ef  and  the  magnitude  and  direction  of  fa  for 
equilibrium. 

Graphic  Method. — Lay  off  the  magnitude-directions  (Fig.  22&) 
from  a  to  e.  The  magnitude-direction  diagram  must  be  closed 
from  e  to  a  by  two  magnitude-directions  ef  and  fa.  Since  the 


FIG.  22a. 


FIG.  22&. 


location-direction  ef  is  vertical,  the  magnitude-direction  ef 
is  also  vertical  and  the  point  /  must  fall  somewhere  on  the  line 
ea,  making  the  magnitude-direction  fa  vertical.  Draw  the 
location-direction  fa.  Choose  any  pole  o  and  draw  the  magni- 
tude-directions 0a,  ob,  oc,  od  and  oe.  Select  any  point  g  on  the 
location-direction  /a;  and  draw  the  location-directions  oa,  ob, 
oCj  od  and  oe',  thereby  locating  the  point  h.  Close  the  diagram 
from  h  to  g  by  drawing  the  location-direction  of.  Draw  the 
magnitude-direction  of  which  determines  the  magnitude-direc- 
tions ef  and  fa.  Scale  the  magnitude-directions  ef  and  /a, 
and  check  by  the  algebraic  method. 

The  magnitude-direction  diagram  becomes  a  straight  line 
in  a  parallel  system. 

In  engineering  practice  the  rigid  body  upon  which  the  forces 


48  THEORY    OF   FRAMED    STRUCTURES  CHAP.  I 

are  acting  is  called  a  beam.  The  forces  acting  downward  are 
known  as  loads  and  the  upward  forces  are  called  reactions. 

3.  The  body  in  Fig.  230,  weighing  1,100  Ib.  and  supported 
at  two  points  X  and  K,  resists  a  horizontal  force  of  350  Ib. 
The  vertical  gravity  line  of  the  body  is  10  ft.  from  the  support 
at  Y.  The  reaction  at  Y  is  650  Ib.  Find  the  direction  of  the 
reaction  at  Y  and  the  magnitude  and  direction  of  the  reaction 
atZ. 

Algebraic  Method.  —  Make  the  sketch  (Fig.  236)  assuming  H- 
and  F-components  for  the  reactions.  Balance  the  moments  of 
the  forces  about  the  left  support  and  thus  eliminate  c  and  d\ 


-  $  =  5,950 

Also  a2  +   b2  =  650* 

whence  a  =  600  or  244.97 

and  b  =  250  or  —  602.07 

Balance  the  H-  and  F-magnitudes 

c  —  600  or  —  252.07 

d  =  500  or  855.03 

The  two  solutions  should  be  indicated  in  separate  sketches,  and 
the  directions  of  the  right  reaction  and  the  magnitudes  and 
directions  of  the  left  reaction  computed  therefrom. 

Graphic  Method.  —  Lay  off  the  known  location-directions  AB 
and  EC  (Fig.  230),  and  the  known  magnitude-directions  ab 
and  be  (Fig.  23^).  The  magnitude-direction  cd  has  a  known 
length  but  an  unknown  direction;  therefore  the  point  d  in  the 
magnitude-direction  diagram  will  fall  on  the  circumference 
drawn  about  c}  with  a  radius  representing  650  Ib.  When  d  is 
located,  the  magnitude-direction  diagram  may  be  closed. 
Through  the  intersection  e  of  the  location-directions  AB  and 
BC,  draw  their  resultant  location-direction  AC.  Only 
one  point  on  each  of  the  location-directions  CD  and  DA  is 
known;  viz.,  the  point  of  each  support.  Draw  the  location- 
direction  OD  connecting  the  points  X  and  F,  and  close  a  loca- 
tion-direction diagram  by  drawing  the  location-directions  OA 
and  OC  intersecting  at  any  point  /  on  the  location-direction 
AC.  Draw  the  corresponding  magnitude-directions  oa  and 
oc  intersecting  at  o.  Through  o  draw  the  magnitude-directions 


SEC.  VI 


EQUILIBRIUM    OF    COPLANAR   FORCES 


49 


od  and  odr  cutting  the  circumference  at  d  and  dr .  The  magni- 
tude-direction diagram  may  now  be  closed  by  drawing  the 
magnitude-directions  cd  and  da,  or  by  drawing  cd!  and  d'a, 
thus  giving  two  solutions. 

For  each  point/  chosen  on  the  location-direction  AC,  there  is 
a  corresponding  pole  o  on  the  line  dd'.     It  was  unnecessary, 


/    B 


FIG.ttb 


except  for  illustration,  to  draw  the  location-direction  AC\  for 
the  location-direction  diagram  could  have  been  closed  at  e  as 
well  as  at  any  other  point/  on  the  location-direction  AC. 

Complete  the  magnitude-direction  diagram,  and  compare 
these  results  with  the  two  algebraic  solutions. 

4.  The  body  in  Fig.  240,  carrying  a  vertical  load  of  15  tons, 
is  supported  by  a  vertical  reaction  of  50  tons;  and  three  other 


THEORY  OF  FRAMED  STRUCTURES 


CHAP.  I 


forces  p,  q  and  s  having  known  locations  and  directions. 
Find  the  magnitudes  p,  q  and  s  if  the  weight  of  the  body  is 
neglected. 

Algebraic  Method. — Make  the  sketch  (Fig.  246)  and  assume 
H-  and  F-components  for  the  inclined  forces  ^>and  q.  The 
unknown  quantities  a,  b  and  5  may  be  determined  by  writing 
and  solving  three  independent  static  equations.  There  is, 
however,  a  very  simple  and  more  direct  method  whereby  any 
one  of  the  unknown  magnitudes  may  be  determined  independ- 


ently  of  the  other  two.  The  method  of  procedure  is  as 
follows:  balance  the  moments  of  all  the  forces  about  the  point 
F,  thereby  eliminating  the  two  unknown  magnitudes  p  and  q 
and  the  load  of  15  tons.  Only  two  magnitudes  remain — the 
reaction  of  50  tons  and  the  magnitude  s — and  their  moments 
about  F  must  balance  for  equilibrium;  for  the  other  magnitudes 
cannot  help  or  hinder  rotation  about  the  point  F.  The 
moment  of  the  reaction  about  F  is  3,000  ft. -tons  clockwise;  and 
the  body  will  rotate  clockwise  about  F  unless  the  magnitude  s, 
acting  through  a  distance  of  40  ft.  and  to  the  right  or  away  from 
the  body,  is  75  tons. 
At  the  point  G,  where  the  directions  of  q  and  s  intersect, 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR   FORCES  51 

resolve  the  magnitude  q  into  H-  and  F-components  30  and  40. 
At  the  point  K  in  a  vertical  line  with  G,  resolve  the  magnitude 
p  into  H-  and  F-components  6b  and  b.  Balance  the  moments 
of  all  the  forces  about  G,  thereby  eliminating  the  unknown 
magnitudes  s,  q  and  b.  The  ^-component  of  p  is  90  tons 

90  X  50  =  4,5oo 
30  X  15  =      45° 
45)4,050 

90  =  6b 
b  =  is 

acting  to  the  left.  The  F-component  of  p  is  15  tons  acting 
downward. 


p  =  \/go2  -\-  i52  =  91.24  tons  acting  toward  the  body. 

The  directions  of  p  and  5  intersect  at  /,  180  ft.  to  the  left  of 
the  reaction;  and  by  balancing  the  moments  of  all  the  forces 
about  /,  the  magnitudes  p,  s  and  30  are  eliminated. 

1 80  X  50  =  9,000 

240  X  15  =  3>6o° 

270)5,400 

20  =  40, 

The  F-component  of  q  is  20  tons  acting  downward.     The  H- 
component  of  q  is  15  tons  acting  to  the  right. 


q  =  Vi52  +  202=25  tons  acting  away  from  the  body. 

The  magnitude  q  may  also  be  determined  by  dividing  the 
moments  of  all  the  forces  about  /  by  the  arm  h.  The  distance 
FG  is  50  ft. 

40  :  50  ::  h  :  270 
h  =  216 

q  =  =  25  tons  acting  away  from  the  body. 

The  magnitudes  3  a  and  40  could  have  been  determined  by 
balancing  the  H-  and  F-magnitudes ;  after  the  magnitudes  s, 
b  and  6b  had  been  determined.  Check  the  solution  already 
determined  by  this  process. 

The  sketch  (Fig.  246)  as  shown  is  incomplete;  since  each 
numerical  value,  as  soon  as  it  is  determined,  should  be  sub- 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  I 


stituted  for  its  symbol  on  the  sketch  and  an  arrow-head  added  to 
indicate  the  sense. 

It  has  been  stated  that  the  use  of  algebraic  equations  should 
be  avoided  whenever  possible.  The  present  case  is  a  good 
illustration.  The  arithmetic  solution  is  not  only  shorter;  but 


FIG.ZSb 


also  gives  the  student  an  opportunity  for  the  exercise  of  his 
judgment  in  choosing  the  best  method  of  attacking  the  problem 
in  order  that  the  solution  may  be  accurately  and  quickly  made, 
and  his  computations  advantageously  arranged  for  checking. 

Graphic  Method. — The  location-directions  of  all  the  forces 
are  known.  They  are  drawn  to  scale  in  Fig.  240  and  indicated 
by  the  symbols  AB,  BC,  CD,  DE  and  EA .  The  known  magni- 
tude-directions ab  and  be  are  laid  off  to  scale  in  Fig.  24^.  Draw 
the  location-directions  OB,  OC  and  OE.  The  forces  acting 


SEC.  VI  EQUILIBRIUM   OF    COPLANAR   FORCES  53 

at  each  of  the  vertices  /,  g  and  h,  of  the  location-direction  dia- 
gram thus  formed,  may  represent  three  concurrent  systems  in 
equilibrium.  Close  the  magnitude-direction  diagram  for  the 
point  /  by  drawing  the  magnitude-directions  co  and  ob,  inter- 
secting at  o.  Two  magnitude-directions  for  the  point  h  are 
now  known — ab  and  bo.  Close  the  diagram  by  drawing  the 
magnitude-directions  oe  and  ea,  intersecting  at  e.  Two  magni- 
tude-directions for  the  point  g  are  known — eo  and  oc.  Close 
the  diagram  by  drawing  the  magnitude-directions  cd  and  ed, 
intersecting  at  d.  The  magnitude-direction  diagram  abcdea 
is  closed  and  the  magnitudes  cd,  de  and  ea  are  determined. 

Compare  the  results  with  those  obtained  by  algebraic  solu- 
tion. 

5.  The  body  in  Fig.  250  supports  three  loads.  What  are  the 
reactions  if  the  weight  of  the  body  is  neglected? 

Algebraic  Method. — Remove  the  body  from  the  supports  and 
indicate  the  H-  and  F-components  of  the  forces  necessary  for 
equilibrium,  as  shown  in  the  sketch  (Fig.  256).  Balance  the 
moments  of  all  the  forces  about  A . 

8  X  210  =     1,680 

16  X  140  =     2,240 

24  X  280  =    6,720 

14)10,640 

760  =  //-component  at  B,  acting  to  the  right. 

=  H  -component  at  A,  acting  to  the  left. 
76°  =  190  =  F-component  at  B  acting  upward. 
4 

210 
140 
280 
630 
IQO 

440  =  F-component  at  A  acting  upward. 
4402  =  193,600 
76o2  =  577,600 

V77i,2oo  =  878.18  =  magnitude  of  the  force  at  A. 
44°   _  ^J     The  direction  of  the  force  at  A  has  the  slope  of  1 1  ver- 
760        19     tical  to  19  horizontal. 
I9O2  =    36,100 
76o2  =  577,600 

Voi3,7oo  =  783.39  =  magnitude  of  the  force  at  B. 


54  THEORY  OF  FRAMED  STRUCTURES        CHAP.  I 

Change  the  location  of  the  components  of  the  lower  reaction 
so  that,  by  balancing  the  moments  of  all  the  forces  about  A  ,  the 
^-component  of  the  lower  reaction  is  eliminated. 

Graphic  Method.  —  The  location-directions  of  the  known  forces 
AB,  BC  and  CD  (Fig.  250)  do  not  intersect;  and  the  conven- 
tional method  of  constructing  the  location-direction  diagram  by 
drawing  the  location-directions  of  the  components  of  the  forces, 
somewhat  after  the  manner  illustrated  in  Problem  2,  would 
ordinarily  be  followed.  But  by  a  simple  expedient  the  location- 
direction  diagram  may  be  more  quickly  drawn  by  locating  the 
resultants  instead  of  the  components.  Draw  the  magnitude- 
direction  diagram  (Fig.  25*;)  from  a  to  d.  Select  two  points 
f  and  g  anywhere  on  the  location-direction  AB  so  that/£  will 
represent  the  magnitude  be.  Similarly  let  hi,  on  the  location- 
direction  BCj  represent  the  magnitude  ab.  The  intersection 
of  the  two  lines  fi  and  gh  at  j  marks  the  location  of  the  resultant 
ac  of  the  two  forces  ab  and  be.  In  like  manner  pq  represents  the 
magnitude  cd\  st  represents  the  magnitude  ac;  and  k  marks 
the  location  of  the  resultant  ad  of  the  forces  ab,  be  and  cd.  The 
location-direction  DE  is  known,  and  intersects  the  location- 
direction  AD  at  m\  through  which  point  the  location-direction 
EA  must  pass  if  there  is  to  be  no  rotation  of  the  body.  The 
location-directions  of  all  the  forces  are  known.  Close  the 
magnitude-direction  diagram  by  drawing  the  magnitude- 
directions  de  and  ea. 

Compare  the  results  with  those  obtained  by  the  algebraic 
solution. 

If  the  sense  of  ab  had  been  upward,  show  that  the  resultant 
of  ab  and  be  would  have  scaled  70  in  the  magnitude-direction 
diagram;  and  its  location  would  have  been  at  the  point  of 
intersection  of  the  two  lines  through  hf  and  ig. 

36.  Problems. 

i.  In  Fig.  26  the  moments  of  all  the  forces  are  balanced  about  the  three 
points  O,  P  and  Q;  i.e. 


o 

ZJIf  o  =  o 

yet  the  body  is  not  in  equilibrium,  for  the  horizontal  magnitudes  do  not  balance, 
neither  do  the  vertical  magnitudes  balance.     Explain. 


SEC.  VI 


EQUILIBRIUM    OF    COPLANAR   FORCES 


55 


2.  The  homogeneous  body  of  uniform  thickness  in  Fig.  27  weighs  50  Ib.  and 
is  supported  at  A,  B  and  C.  The  reaction  at  A  is  25  Ib.  The  reactions  at  B 
and  C  are  vertical  and  horizontal  respectively,  as  indicated  by  the  roller  supports. 
Determine  the  unknown  elements  for  equilibrium. 


3.  The  body  in  Fig.  28  is  supported  at  A,  B  and  C.  Find  the  weight  of  the 
body,  the  magnitude  of  the  force  at  A}  and  locate  the  vertical  gravity  line  of  the 
body. 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  I 


4.  The  cross-section  of  a  retaining  wall  weighing  6,000  Ib.  per  linear  foot  is; 
represented  in  Fig.  29.     The  resultant  earth  pressures  per  linear  foot  are  15,000 
Ib.  at  A  and  7,500  Ib.  at  B.     What  is  the  direction  of  the  resultant  pressure  at 
A,  if  the  resultant  reaction  of  the  ground  on  the  base  CD  is  25,500  Ib.  per  linear 
foot;  and  what  is  the  total  friction  between  the  ground  and  wall  on  the  line  CD? 
At  what  point  on  the  line  CD  does  the  resultant  reaction  act? 

5.  Neglect  the  weight  of  the  body  in  Fig.  30  and  determine  the  reactions. 

6.  The  body  in  Fig.  31  weighs  2,000  Ib.    Locate  the  vertical  gravity  line; 


find  the  magnitude  of  the  force  at  A;  and  the  direction  of  the  force  of  1,300  Ib. 
atB. 

7.  Find  the  weight  of  the  body  in  Fig.  32;  the  magnitude  of  the  force  at  A; 
and  locate  the  force  of  1,000  Ib.  for  equilibrium. 

8.  How  far  is  the  ball  (Fig.  33)  from  the  point  A  when  the  body  is  in  equilib- 
rium; what  is  the  magnitude  and  direction  of  the  force  at  A?    Neglect  the 
weight  of  the  body. 

9.  The  body  in  Fig.  34  weighs  1,000  Ib.  and  resists  two  forces  at  C  as  indi- 
cated.    The  reactions  at  A  and  B  are  820  Ib.  and  1,130  Ib.  respectively.     In 
what  direction  does  each  reaction  act  and  where  is  the  vertical  gravity  line  of 
the  body? 


SEC.  VI  EQUILIBRIUM    OF    COPLANAR   FORCES 


57 


FIG.  33- 


^ 22'- 


45' 


. 

40* 


60 


95* 


FIG.  35- 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  I 


10.  Neglect  the  weight  of  the  body  (Fig.  35)  and  find  the  magnitudes  of  the 
forces  p,  q  and  s. 


400* 


16' 


FIG.  36. 


FIG.  37. 


ii.  Find  the  weight  and  vertical  gravity  line  of  the  body  (Fig.  36)  if  the 
magnitude  of  the  force  at  A  is  781  Ib.     What  is  the  direction  of  the  force  at  A  ? 

12.  Neglect    the    weight    of    the 
body  (Fig.  37)  and  find  the  magni- 
tude of  A,  and    the    location    and 
direction  of  a  force  of  1,220  Ib.  for 
equilibrium. 

13.  A  level  beam  of  uniform  cross- 
section  10  ft.  long,  weighing  250  Ib. 
and  supported  at  each  end,  carries  a 
single  load  of  5   tons  at  mid-span. 
Why  is  the  problem  of  finding  the 
two  reactions  indeterminate? 

An  algebraic  solution  for  each  of 
the  two  following  problems  is  long 


FIG.  38. 

and    involved. 

graphic  solution  is  comparatively  short  in  each  case. 


Show    that    the 


FIG.  39. 


SEC.  VI  EQUILIBRIUM   OF   COPLANAR   FORCES  59 

14.  The  magnitudes  of  the  forces  at  A  and  B  (Fig.  38)  are  13  Ib.  and  17  Ib. 
respectively.     Neglect  the  weight  of  the  body  and  find  the  directions  of   the 
forces  at  A  and  B,  and  the  magnitude  of  the  force  C. 

15.  The  magnitudes  of  the  forces  at  A,  B  and  C  (Fig.  39)  are  41  Ib.,  50  Ib.  and 
25  Ib.  respectively.     Neglect  the  weight  of  the  body  and  find  the  three  unknown 
directions. 


CHAPTER  II 
APPLICATION  OF  THE  PRINCIPLES  OF  EQUILIBRIUM 

SEC.  I.     SIMPLE  TRUSSES 

37.  Rigid  Body. — In  the  foregoing  section  reference  is 
frequently  made  to  a  "rigid  body."  No  such  thing  exists  in 
nature,  if  "rigid"  is  used  in  the  sense  of  "unyielding."  All 
solids  may  be  considered  to  approximate  in  a  greater  or  less 
degree  an  ideal  state  of  complete  rigidity.  A  steel  casting,  a 
granite  rock,  an  oak  block,  putty  and  rubber — all  illustrate 
different  degrees  of  rigidity,  for  all  change  their  shape  under 
different  degrees  of  pressure.  In  engineering  parlance  a  rigid 
body  is  one  which,  in  form  and  content,  offers  a  sufficient 
resistance  to  distortion  to  give  service  as  an  engineering  struc- 
ture. Or,  in  other  words,  if  the  body  is  a  solid,  as  an  oak  beam 
or  steel-plate  girder,  consideration  is  given  to  the  quality  and 
quantity  of  the  material  in  order  that  a  sufficient  degree  of 
rigidity  may  be  realized;  but  if  the  body  is  a  framed  structure, 
attention  must  also  be  given  to  the  manner  in  which  the 
structure  is  framed  in  order  that  it  may  be  stable,  or  not  be 
susceptible  to  an  immediate  and  complete  collapse.  The  frame 
represented  in  Fig.  2ih  is  unstable  or  collapsible;  for  it  offers 
no  resistance  to  a  change  in  its  shape,  except  possibly  a  small 
amount  of  friction  at  the  joints.  The  shape  of  the  frame  may 
be  changed  at  will  without  affecting  a  change  in  the  length  of 
any  member,  and  the  same  is  true  of  any  similar  frame  having 
four  or  more  members. 

A  triangular  frame  differs  from  all  others  in  this  respect, 
since  a  change  in  its  shape  necessitates  a  change  in  the  length  of 
at  least  one  of  its  three  members.  Hence  a  frame,  which  in 
outline  presents  a  collection  of  triangles,  is  stable;  and  if  it  has 
adequate  material  in  its  members,  it  possesses  a  sufficient 
degree  of  rigidity  for  engineering  purposes.  Hence: 

60 


SEC.  I          APPLICATION   OF   PRINCIPLES    OF   EQUILIBRIUM  6 1 

38.  A  Truss  is  a  frame  or  jointed   structure  composed  of 
members  (usually  straight)  so  connected  as  to  form  a  succession 
of  triangles  making  one  rigid  or  stable  structure.     Acting  as  a 
whole,  this  performs  the  functions  of  a  beam  in  resisting  dis- 
tortion1 caused  by  shearing  forces  and  bending  moment;  while 
the  individual  members  are  designed  (except  in  special  cases)  to 
perform  the  functions  of  a  tie  or  strut  in  resisting  changes  in 
their  length  caused  by  tensile  or  compressive  forces. 

39.  Stability. — The  simplest  truss  that  can  be  constructed 
has  three  members  m,  and  three  joints  j.     A  more  elaborate 
truss  comprises  several  triangular  frames,  so  combined  that  each 
additional  triangle  adds  one  joint  and  two  members.     Hence, 
the  number  of  members  necessary  to  insure  stability  under  any 
arrangement  of  loading  is 

m  =  y  -  3  (i) 

A  truss  having  fewer  members  than  are  required  by  Eq.  (i)  is 
in  a  state  of  unstable  equilibrium,  and  will  collapse  except 
under  special  conditions  of  loading.  A  truss  having  more 
members  than  are  indicated  by  Eq.  (i)  is  a  redundant  structure. 
Such  structures,  if  the  members  are  properly  placed,  will 
support  loads  of  any  arrangement.  In  some  instances  this  is  a 
distinct  advantage.  Redundant  structures  cannot  be  analyzed 
by  the  principles  of  statics  alone,  and  are  said  to  be  "  statically 
indeterminate."  By  giving  due  consideration  to  the  elastic 
properties  of  the  members;  or  by  the  aid  of  certain  reasonable 
assumptions  which  have  the  sanction  of  good  engineering 
practice,  a  satisfactory  analysis  may  be  made. 

40.  The  assumption  that  the  members  of  a  truss  are  subject 
to    tensile    and    compressive    forces    only,    presupposes    four 
conditions : 

]  It  should  be  clearly  understood  and  constantly  kept  in  mind  that  structures 
are  not  rigid  in  the  sense  of  unyielding,  but  are  elastic  in  the  same  sense  that  a 
rubber  eraser  or  a  rubber  band  is  elastic,  only  in  a  lesser  degree.  The  members 
of  a  steel  truss,  designed  in  accordance  with  current  specifications,  will  lengthen 
or  shorten  from  3^2  to  ^  e  of  an  inch  for  every  10  ft.  of  length  when  the  structure 
is  supporting  the  full  load  for  which  it  was  designed.  This  change  in  length 
inevitably  changes  the  shape  of  the  triangular  units  of  the  truss  and  accounts  for 
the  distortion  of  the  whole  structure.  The  distortion  is  comparatively  so  slight, 
however,  that  for  our  present  consideration,  the  changes  in  the  dimensions  of  the 
truss  may  be  considered  negligible  without  appreciable  error. 


62  THEORY    OF   FRAMED    STRUCTURES  CHAP.  II 

1.  That  the  axial    gravity  lines  of  the  members  meeting 
at  a  joint,  intersect  at  one  point. 

2.  That  the  loads  are  applied  only  at  the  joints. 

3.  That  the  members  are  subject  to  neither  shear  nor  bending 
moment  due  to  their  own  weight. 

4.  That  the  joints  are  frictionless  hinges. 

The  first  and  second  conditions  are  generally  incorporated 
in  a  good  design.  The  third  condition  is  true  of  vertical  mem- 
bers only.  The  nearest  approximation  to  the  fourth  condition 
is  found  in  a  pin-connected  truss. 

41.  The  application  of  the  principles  of  equilibrium  to  the 
analysis  of  a  framed  structure  may  be  considered  as  two  distinct 
operations:  (a)  the  determination  of  the  external  forces;  and  (b) 
the  determination  of  the  internal  forces. 

42.  The  external  forces  acting  on  a  structure  are:  (a)  the 
loads  which  the  structure  supports;  (b)  the  weight  of  the  struc- 
ture itself;  and  (c)  the  forces  acting  at  the  points  of  support, 
commonly  called  the  reactions. 

The  loads  are  usually  known  or  easily  determined,  but  the 
weight  of  the  structure  not  as  yet  designed  must  be  assumed. 
The  reactions  are  determined  by  a  solution  of  a  system  of  non- 
concurrent  forces,  as  set  forth  in  Sections  V  and  VI  of  Chapter  I. 
The  external  forces  are  generally  analyzed  by  the  algebraic 
method.  This  process  is  usually  shorter  than  the  graphic 
method,  especially  if  the  external  forces  are  parallel. 

43.  The  internal  forces  acting  in  the  structure  are  many  and 
of  various  kinds.     We  shall  consider  here  only  those  of  primary 
importance — the  tensile  and  compressive  forces  acting  along 
the  members.     These  forces  are  known  as  stresses. 

As  the  reactions  are  determined  by  the  application  of  the 
principles  of  the  equilibrium  of  coplanar  forces  to  the  structure 
as  a  whole,  so  the  stresses  in  the  members  are  determined  by  the 
same  principles;  but  applied  to  isolated  parts  of  the  structure. 
The  portion  to  be  considered  may  be  a  single  joint,  or  it  may 
include  several  joints  and  members.  The  forces  to  be  con- 
sidered may  represent  reactions,  loads  or  stresses.  There  are 
two  principal  methods  of  procedure  in  the  determination  of 
stresses  in  framed  structures: 


SEC.  I         APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM  63 

1.  The  method  of  joints 

(a)  algebraic 

(b)  graphic. 

2.  The  method  of  sections 

(a)  algebraic 

(b)  graphic. 

44.  The  Method  of  Joints. — The  truss  in  Fig.  400  supports 
five  loads  as  shown.  What  is  the  stress  in  each  member  if  the 
reactions  are  vertical? 

The  determination  of  the  stresses  by  the  method  of  joints, 
either  algebraically  or  graphically,  is  simply  a  solution  of  as 
many  systems  of  concurrent  forces  as  there  are  joints  in  the 
truss.  After  the  external  forces  have  been  found,  the  solution 
begins  at  a  joint  where  the  stresses  of  two  members  concur  with 
one  or  more  external  forces;  proceeds  to  an  adjacent  joint  which 
presents  but  two  unknown  stresses;  and  so  on  throughout  the 
entire  structure.  The  force  exerted  on  a  joint  by  one  end  of 
a  member  is  equal  in  magnitude  but  opposite  in  sense  to  the 
force  exerted  by  the  member  on  the  joint  at  its  other  end. 
The  kind  of  stress  in  a  member  is  indicated  by  the  sense  of  the 
force  which  a  member  exerts  on  a  joint;  i.e.,  the  stress  is  tensile 
if  the  sense  is  away  from  the  joint  and  compressive  if  toward 
the  joint. 

Graphic  Solution. — The  first  step  in  the  solution  of  any  prob- 
lem of  this  character  is  the  determination  of  the  forces  necessary 
to  place  the  truss  in  equilibrium,  i.e.,  to  find  the  reactions.  The 
left  and  right  reactions,  determined  algebraically,  are  50  Ib.  and 
55  Ib.  respectively.  It  is  a  waste  of  time  to  find  the  reactions 
graphically  when  the  external  forces  are  parallel.  A  capital 
letter  is  placed  in  the  space  between  each  external  force  and  in 
each  triangle  of  the  frame.  The  outline  of  the  truss  represents 
essentially  a  location-direction  diagram,  and  we  proceed  to 
draw  a  closed  magnitude-direction  diagram  for  each  of  the  six 
concurrent  systems — one  for  each  joint.  Since  the  end  joints 
present  but  two  unknowns,  either  may  be  chosen  for  the  draw- 
ing of  the  first  diagram.  The  conventional  method  is  to  begin 
at  the  left  end  of  a  structure,  proceeding  in  a  clockwise  rotation 
about  each  joint.  All  location-directions  are  known.  The 


64 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  II 


problem  is  to  determine  unknown  magnitudes  or  stresses  acting 
in  the  various  members. 

Beginning  at  the  left-end  joint,  proceeding  in  a  clockwise  rota- 
tion, and  including  first  the  location-directions  having  known 


K50  +9$  0 

(a)     Scale  fain  -5 ft 


•f  =  Tension 
a          -  =  Compression 


(9) 


FIG.  40. 


magnitudes,  we  read  GA,  AB,  BE  and  EG.  The  magnitude- 
directions  ga,  abj  bh  and  hg  are  laid  off  in  the  same  order  in  Fig. 
406.  The  magnitude-directions  ga  and  ab  are  known,  and  the 
diagram  is  closed  by  drawing  bh  and  hg.  The  sense  bh  is  toward 
the  joint;  the  sense  hg  is  away  from  it.  The  magnitude-direc- 
tion bh  scales  104  lb.;  the  magnitude-direction  hg  scales  96  lb.; 


SEC.  I         APPLICATION   OF   PRINCIPLES   OF  EQUILIBRIUM  65 

hence  the  stresses  in  BH  and  EG  are  104  Ib.  compress! ve,  and 
96  Ib.  tensile  respectively.  These  numerical  values,  with  minus 
and  plus  signs  to  signify  compression  and  tension,  are  indicated 
on  the  members  of  the  truss. 

The  adjacent  upper  joint  now  presents  only  two  unknown 
stresses.  The  known  magnitude-directions  hb  and  be  are  laid 
off  (Fig.  4oc)  and  the  diagram  closed  by  drawing  cj  and  jh\ 
giving  —78  Ib.  and  —  26  Ib.  for  the  stresses  in  the  members  CJ 
and  JH. 

At  the  lower  middle  joint  the  stresses  in  three  members  JK, 
KL  and  LG  are  unknown  and  a  solution  is  not  yet  possible; 
but  at  the  upper  middle  joint  the  stresses  in  only  two  members 
are  unknown,  and  the  solution  is  given  in  Fig.  40^. 

There  are  but  two  unknowns  at  each  of  the  three  remaining 
joints.  Fig.  406  represents  the  diagram  for  the  lower  middle 
joint,  and  Fig.  4o/,  the  joint  supporting  the  load  of  40  Ib.  The 
stresses  in  all  the  members  have  been  determined,  and  all 
the  forces  concurring  at  the  right-end  joint  are  known ;  but  the 
diagram  (Fig.  40^)  is  drawn  for  this  joint  as  a  check  on  the 
foregoing  solutions. 

Any  two  adjacent  joints  have  one  member  in  common,  con- 
sequently their  respective  magnitude-direction  diagrams  also 
have  one  side  in  common;  hence,  any  one  of  the  diagrams  may 
be  added  to  the  one  preceding  or  following  it.  Thus  one  figure 
may  be  developed  which  will  contain  all  the  magnitude-direc- 
tion diagrams  representing  graphically  the  magnitude,  direction 
and  sense  of  each  external  force  and  internal  stress  of  the  struc- 
ture. Such  a  figure  is  called  a  stress  diagram  and  is  illustrated 
in  Fig.  4oh.  The  magnitude-direction  diagram  of  the  external 
forces  or  load  line  ab,  be,  cd,  de,  ef,  fg  and  ga  is  drawn  first  and 
the  lines  representing  the  stresses  follow  in  regular  order.  Be- 
ginning at  the  left-end  joint,  we  read  ga,  ab  and  draw  bh  and  kg 
parallel  respectively  to  BH  and  EG.  For  the  next  joint  we 
read  hb,  be  and  draw  cj  and  jh.  For  the  top  joint  we  read  jc, 
cd  and  draw  dk  and  kj.  For  the  bottom  joint  we  read  gh,  hj 
and  jk,  and  draw  kl  and  Ig.  For  the  next  joint  we  read  Ik,  kd 
and  de  and  draw  el  parallel  to  EL.  If  this  line,  drawn  from  e 
parallel  to  EL  passes  through  the  point  /,  we  have  a  check 


66 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  II 


on  our  work,  the  stress  diagram  closes  and  the  solution  is 
complete. 

When  the  external  forces  are  parallel,  as  in  the  present  case, 
the  magnitude-direction  diagram  for  the  external  forces  is  a 
straight  line.  The  stress  diagram  is  essentially  a  magnitude- 
direction  diagram  for  the  external  forces  and  a  magnitude- 
direction  diagram  for  the  concurrent  system  at  each  joint.  ' 

Algebraic  Solution.—  The  concurrent  systems  are  sketched  in 
Fig.  41.  The  F-component  of  a  (Fig.  a)  is  40  Ib.  acting  down- 


30 


FIG.  41. 
I  2 

ward.     The  //-component  is  -  -  X  40  =  96  Ib.  acting  to  the 

o 

left;  hence,  b  =  96  Ib.  acting  to  the  right  or  away  from  the  joint; 
a  =  —  X  40  =  104  Ib.  acting  toward  the  joint.  Assume  //- 
and  F-components  for  c  and  d  as  shown  in  Fig.  b. 

5^+20    =  5;y  +  40 
120;  +  123;  =  96 


x  =  6 
y  =  2  ^ 

//-component  of  c  =  72  Ib.  acting  to  the  left. 
F-component  of  c  =  30  Ib.  acting  downward. 

c  =  78  Ib.  acting  toward  the  joint, 
//-component  of  d  =  24  Ib.  acting  to  the  left. 
F-component  of  d  =  10  Ib.  acting  upward. 

d—  26  Ib.  acting  toward  the  joint. 

The  upper  middle  joint  is  sketched  in  Fig.  c.     The  //-com- 
ponent of  c  =  72  Ib.  acting  to  the  right,  consequently  the  H 


SEC.  I       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM  67 

component  of/  =  72  Ib.  acting  to  the  left.  Hence,  the  F-com- 
ponent  of/  =  30  Ib.  acting  upward;/  =  78  Ib.  acting  toward  the 
joint;  and  e  =  30  Ib.  acting  downward  or  away  from  the  joint, 
to  balance  the  load  of  30  Ib.  and  the  F-components  of  c  and/. 

At  the  lower  middle  joint  (Fig.  d)  the  F-component  of  g  = 
20  Ib.,  acting  downwardt  o  balance  e  and  the  F-component  of 
d;  therefore  the  //-component  of  g  =  48  Ib.  acting  to  the  left, 
and  g  =  52  Ib.  acting  toward  the  joint.  Hence,  k  =  120  Ib., 
acting  away  from  the  joint. 

In  Fig.  e  there  is  but  one  unknown  force/.  The  //-compo- 
nent =  120  Ib.  acting  to  the  left,  the  F-component  =  50  Ib. 
acting  upward  and  j  =  130  Ib.  acting  toward  the  joint.  We 
now  have  a  check  upon  our  computations  by  noting  that  the 
F-component  of  j  equals  the  right  reaction,  minus  the  load  of 
5  Ib.;  the  //-component  equals  &;  and  the  ratio  of  the  F-  and 

CQ  C 

//-components  of  j  is  -!L-  or  —  >  which  is  the  slope  or  bevel  of 
120       12 

the  member. 

Assuming  that  the  stresses  in  /?//  and  CJ  are  unknown,  solve 
algebraically  and  graphically  for  the  stress  in  ///,  using  the 
method  of  joints. 

Remove  the  loads  of  20  Ib.  and  40  Ib.,  and  determine  the 
stresses  in  the  three  web  members  ///,  JK  and  KL. 

The  graphic  solution  is  generally  employed  when  the  stresses 
are  determined  by  the  method  of  joints.  This  is  especially 
true  in  the  case  of  roof  trusses  having  inclined  chords.  There  is 
perhaps  little  preference  between  the  algebraic  and  the  graphic 
solutions,  if  the  top  and  bottom  chords  are  horizontal  and  half 
the  web  members  are  vertical.  There  are  several  factors 
which  conspire  in  making  the  method  of  joints  peculiarly 
efficient  in  truss  analysis  for  stationary  loads. 

1.  The  stress  in  any  member  is  constant  for  a  given  load. 

2.  The  required  stresses  in  all  members  occur  simultaneously. 

3.  The  external  forces  and  required  stresses  concurring  at  a 
joint  are  in  equilibrium. 

The  stress  in  any  member  of  a  truss  supporting  moving  loads 
varies  as  the  loads  move  upon  the  structure.  The  stress  in  any 
member  is  desired  only  for  that  particular  position  of  the  loads 


68  THEORY    OF   FRAMED    STRUCTURES  CHAP.  II 

which  will  cause  the  maximum  stress  in  that  member.  Some 
members  receive  their  maximum  stress  when  the  moving  loads 
cover  the  entire  structure,  while  in  other  members  the  stress  is 
greatest  when  only  a  portion  of  the  structure  is  covered.  The 
maximum  stresses  do  not  occur  simultaneously.  This  fact  greatly 
limits  the  usefulness  of  the  stress  diagram  in  the  case  of  moving 
loads.  For  example,  suppose  that  four  members  A,  B,  C  and 
D  meet  at  a  joint;  and  that  the  moving  loads  are  in  the  position 
for  maximum  stress  in  B.  A  stress  diagram  will  give  the 
stresses  not  only  in  £;  but  in  A,  C,  D  and  all  the  other  members 
of  the  structure  as  well.  The  stress  in  B,  however,  is  the  only 
stress  of  any  value  which  may  be  obtained  from  that  particular 
stress  diagram;  since  it  is  the  only  member  in  which  the  stress 
is  a  maximum. 

45.  The  method  of  sections  gives  excellent  service  when  the 
stress  in  but  one  member  is  desired  for  any  particular  arrange- 
ment of  loads.  The  algebraic  solution  is  generally  preferred. 
Suppose  we  wish  to  determine  the  stress  in  the  member  GH 
(Fig.  420).  If  the  member  were  cut  or  removed  from  the  truss, 
it  is  perfectly  obvious  that  the  part  ACG  would  rotate  clock- 
wise, and  the  part  CMH  would  rotate  counter-clockwise — each 
about  the  pivot  C;  the  joints  G  and  H  moving  further  apart. 
Hence  in  preserving  the  equilibrium  of  the  structure,  it  is 
plainly  the  duty  of  the  member  GH  to  hold  the  joint  G  from 
moving  to  the  left,  and  the  joint  H  from  moving  to  the  right. 
In  performing  this  function  the  member  exerts  a  force  to  the 
right  on  the  joint  G,  and  an  equal  and  opposite  force  to  the  left 
on  the  joint  H. 

Imagine  that  a  plane  of  section  XY  cuts  the  three  members 
CD,  CH  and  GH,  dividing  the  truss  into  two  portions.  If  p,  q 
and  s  represent  the  stresses  in  the  members  before  cutting, 
then  the  known  external  forces  (loads  and  reactions)  acting 
upon  each  portion  are  balanced  by  the  forces  p,  q  and  s  as 
illustrated  in  Figs.  426  and  c.  The  left  portion  (Fig.  426) 
represents  a  coplanar  system  of  five  non-concurrent  forces 
having  three  unknown  magnitudes;  corresponding  to  the  first 
combination  listed  in  Article  34.  An  algebraic  and  graphic 
solution  for  this  system  is  given  in  Article  35,  Problem  4. 


SEC.  I         APPLICATION   OF   PRINCIPLES    OF   EQUILIBRIUM 


69 


The  algebraic  solution  when  performed  by  writing  and 
solving  three  simultaneous  equations  is  known  as  Rankine's1 
"Method  of  Sections."  The  corresponding  graphic  solution 
is  the  work  of  Culmann.2  The  algebraic  solution,  whereby 
any  one  of  the  three  unknown  magnitudes  is  determined  by 
balancing  the  moments  of  all  the  forces  about  the  intersection 
of  the  other  two  forces  having  unknown  magnitudes,  is  known 
as  Ritter's3  "  Methods  of  Moments."  In  order  to  determine  the 
stress  by  Ritter's  method,  we  balance  the  moments  of  all  the 
external  forces  about  C  (Figs.  426  or  c)  thereby  eliminating  the 
two  unknown  magnitudes  p  and  q. 


I      D 


V  V    y 


FIG.42C 


Moments  about  C 

FIG.  426 
60  X  50  =  3,000 


20 


40 


30 


10 


Fl  6.43d 


30  X    8  = 

60  X  20  = 

90  X  40  = 

120  X  30  = 

150  X  10  = 


FIG.  420 

240     1 80  X  73 

1,200 
3,600 
3,600 
1,500 


13,140 
10,140 

3,ooo 


10,140 


1  Applied  Mechanics. 

2  Die  Graphische  Statik.     1886. 

3  Dach-  und  Brucken  constructionen. 


1862. 


70  THEORY    OF   FRAMED    STRUCTURES  CHAP.  II 

The  sum  of  the  moments  about  C  of  the  known  forces  acting 
on  the  left-hand  portion  is  3,000  ft. -tons  clockwise;  hence,  the 
force  5  acting  at  a  distance  of  40  ft.  from  C  must  have  a  magni- 
tude of  75  tons  toward  the  right  or  away  from  the  joint  G  to 
balance  the  moments.  The  sum  of  the  moments  about  C  of 
the  known  forces  acting  on  the  right-hand  portion  is  also  3,000 
ft.-tons,  but  counter-clockwise;  hence,  the  force  s  acting  at  a 
distance  of  40  ft.  from  C  must  have  a  magnitude  of  75  tons 
toward  the  left,  or  away  from  the  joint  H  to  balance  the 
moments. 

The  fact  that  the  sum  of  the  moments  about  C  of  the  known 
external  forces  is  the  same  (3,000  ft.-tons)  for  either  portion, 
except  that  the  rotation  is  clockwise  in  one  case  and  counter- 
clockwise in  the  other,  is  easily  explained.  The  truss  (Fig. 
42 a)  is  in  equilibrium  and  the  sum  of  the  moments  of  the  eight 
known  external  forces  about  any  point  C,  for  example,  equals 
zero.  If  these  eight  forces  are  divided  in  any  manner  into  two 
groups,  the  sum  of  the  moments  of  one  group  about  C  must 
balance  the  sum  of  the  moments  of  the  other  group. 

Determine  p  and  q  (Fig.  420)  by  balancing  the  moments  of 
the  forces  about  the  points  H  and  /  respectively,  and  note  that 
the  results  check  with  the  computations  of  Problem  4,  Article 

35- 

Since  the  unknown  magnitudes  may  be  determined  by 
balancing  the  moments  of  the  forces  on  either  side  of  the  section, 
it  is  expedient  to  consider  the  side  which  has  the  fewer  forces. 

The  stress  in  DJ  (Fig.  420)  may  be  determined  by  passing  a 
section  through  DE,  DJ  and  HJ.  The  two  chord  members  are 
parallel  and  there  is  no  point  of  their  intersection  about  which 
the  moments  may  be  balanced;  but  since  the  chords  are  hori- 
zontal, the  vertical  component  of  the  stress  in  DJ  must  balance 
the  vertical  magnitudes  on  either  side  of  the  section.  The 
resultant  of  the  vertical  forces  on  the  left  of  the  section  is 
50—  (15  +  8)  =  27  tons  acting  upward.  The  left-hand 
portion  will  move  upward,  unless  the  vertical  component  of 
DJ  is  27  tons  acting  downward  from  D  or  away  from  the 
joint.  Therefore  the  member  is  in  tension  and  the  vertical 
component  of  the  stress  is  27  tons. 


SEC.  I         APPLICATION   OF   PRINCIPLES    OF   EQUILIBRIUM 


71 


Compute  the  stresses  in  DE  and  HJ  and  see  if  these  stresses 
balance  with  the  horizontal  component  of  DJ. 

Compute  the  stresses  in  BCj  BG,  CG  and  DH  and  then  draw  a 
stress  diagram  to  check  the  computations. 

It  is  worthy  of  note  that  a  solution  by  the  method  of  sections 
is  entirely  independent  of  the  number,  inclination  or  arrange- 
ment of  any  members  of  the  structure  other  than  those  cut  by 
the  section. 

46.  Problems. 

i.  The  truss  in  Fig.  43  is  supported  by  vertical  reactions  at  each  end.     Deter- 


400 


500 


•  6  Equal  Spaces  * 

FIG.  43. 

140 


280 


FIG.  44. 

mine  the  reactions  algebraically,  construct  the  magnitude-direction  diagram  for 
the  external  forces  (sometimes  called  the  load  line)  and  draw  a  stress  diagram. 
Mark  the  stresses  with  proper  signs  upon  the  members.  Note  that  h  and  i  fall 
at  the  same  point  in  the  stress  diagram,  indicating  that  the  member  HI  has  no 
stress  and  is  superfluous  for  this  particular  loading.  Note  also  that  the  stresses 
in  QF  and  GP  are  equal,  and  that  the  stress  in  PQ  equals  the  external  load  at 
the  joint.  In  consideration  of  these  observations  develop  a  general  rule  by 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  II 


which  superfluous  members  may  be  eliminated  before  the  stress  diagram  is 
begun. 

2.  The  frame  in  Fig.  44  has  been  substituted  for  the  solid  body  of  Fig.  2$a. 
Draw  a  stress  diagram.     Note  that  the  magnitude-direction    diagram  for  the 
external  forces,  which  constitutes  the  beginning  or  foundation  of  the  stress 
diagram,  has  already  been  constructed  in  Fig.  25^. 

3.  Substitute  a  frame  for  the  solid  body  in  Fig.  30  and  draw  a  stress  diagram. 

4.  The  truss  in  Fig.  45  is  supported  by  vertical  reactions  at  A  and  B.     Draw 
a  stress  diagram.     Compute  the  stresses  in  X ,  Y  and  Z. 


2QO 


400 


1000 


FIG.  45. 


3000 


F1G.49 

5.  Compute  the  H-  and  F-components  of  the  stresses  in  the  five  members 
meeting  at  A  (Fig.  46).     Mark  the  values  on  a  sketch  and  indicate  whether  the 
members  are  in  tension  or  compression. 

6.  The  truss  in  Fig.  47  is  supported  at  A  and  B.     The  roller  at  B  indicates 
a  vertical  reaction.     Draw  a  stress  diagram. 

7.  The  truss  in  Fig.  48  is  supported  by  a  chain  at  D.     Suspend  weights  of 
150  Ib.  at  A,  50  Ib.  at  B  and  50  Ib.  at  C,  and  draw  a  stress  diagram. 

8.  The  truss  in  Fig.  49  is  supported  by  chains  AC  and  BD.     Draw  a  stress 
diagram  for  vertical  loads  of  300  Ib.  at  E\  600  Ib.  at  F;  and  a  horizontal  load  of 
450  Ib.  at  G,  acting  to  the  right. 


SEC.  II       APPLICATION   OF   PRINCIPLES   OF   EQUILIBRIUM  73 

SEC.  II.     STRUCTURES  REQUIRING  SPECIAL  CONSIDERATION 

47.  The  application  of  the  principles  of  static  equilibrium 
to  the  framed  structures,  discussed  in  the  preceding  section, 
presented  no  real  difficulties.     There  are  types  of  structures, 
however,  for  which  the  solutions  are  not  so  simple;  and  a 
special  investigation  is  necessary  in  order  that  the  difficulties 
encountered  may  be  overcome.     These  difficulties  are  caused 
by  indeterminate  reactions,  indeterminate  stresses   or  both. 
Statically  indeterminate  structures  as  such  will  not  be  treated 
here.     There  are,  however,  several  indeterminate  types  which 
may  be  treated  by  static  methods;  when  certain  reasonable 
assumptions   which   have   the   sanction   of   good   engineering 
practice  are  made.     There  are  also  several  types  which  are 
indeterminate  in  appearance  but  not  in  fact. 

48.  The  Fink  Truss. — The  structure  outlined  in  Fig.  500, 
and  known  in  this  country  as  a  Fink  truss,  is  a  good  example  of  a 
case  in  which  the  stresses  are  apparently  statically  indeter- 
minate.    A  difficulty  is  encountered  after  the  stress  diagram 
(Fig.  506)  has  been  drawn  for  joints  i,  2  and  3;  for  the  stresses 
in  three  members  at  each  of  the  joints  4  and  5  are  unknown. 
In  this  emergency,  one  of  two  graphic  solutions  (a)  or  (b)  may 
be  used. 

(a)  Temporarily  remove  the  members  KL  and  LM ,  substitute 
the  dotted  member  OM  connecting  the  joints  5  and  7,  and  draw 
the  polygon  jhdeof  for  the  joint  4.     Then  the  polygon  oefmo  for 
joint  7  may  be  drawn,  giving  the  stress  in  the  member  FM ; 
which  is  in  no  way  influenced  by  the  arrangement  of  members 
in  the  quadrilateral  4-5-6-7.     Remove  the  dotted  member  and 
replace  the  members  KL  and  LM.    The  point  m  is  now  located 
in  the  stress  diagram;  the  stress  in  FM  is  known;  and  the 
remaining  stresses  are  easily  determined  by  taking  the  joints 
in  the  order  7-4-5-6. 

(b)  The  following  solution  is  based  upon  the  exception  to  the 
general  law  of  concurrent  forces  laid  down  in  Article  17.     The 
point  j  was  located  in  the  stress  diagram  when  joint  3  was 
solved.     The  point  k  will  fall  somewhere  on  the  line  through  j 
parallel  to  KJ.     The  stress  in  ML  can  be  determined  since  the 


74 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  II 


members  LE  and  FM  are  in  the  same  straight  line.  Assume 
any  value  for  the  stress  in  the  member  LE,  as  l'e\  and  draw  the 
stress  polygon  I'efm'l',  giving  m'l'  for  the  stress  in  ML.  Since 
there  remain  but  three  members  having  unknown  stresses  at 
joint  6,  and  two  of  the  members  are  in  the  same  straight  line, 
the  stress  polygon  I'm'n'k'l'  can  be  drawn  and  the  stress  in  KL 
determined. 

Any  other  value  I'e  might  have  been  assumed  for  the  stress  in 
the  member  LE\  and  for  each  different  point  /'  on  the  line 


FIG.SOf 


through  a  parallel  to  LE,  there  is  a  corresponding  point  kf  on  a 
line  parallel  to  LE.  Hence  the  point  k  is  located  at  the  inter- 
section of  the  line  through  j  parallel  to  KJ,  with  the  line 
through  kf  parallel  to  LE.  The  stresses  can  now  be  solved  for 
either  joint  4  or  5  and  the  diagram  completed  without  any 
further  difficulty. 

Complete   the   stress   diagram   for  solutions    (a)    and    (b). 

When  computing  the  stresses  by  the  method  of  sections,  we 
meet  with  a  difficulty  somewhat  similar  to  the  one  encountered 
in  drawing  the  stress  diagram.  While  it  is  impossible  to  divide 
the  truss  by  a  section  cutting  any  one  of  the  members  JK,  KN, 
KL,  LE  or  ML  without  cutting  more  than  three  members; 
yet  the  stresses  in  these  members  are  easily  determined  when 


SEC.  II       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM  75 

taken  in  the  proper  order.  The  stress  in  the  member  ML  may 
be  determined  by  a  section  cutting  out  the  portion  shown  in 
Fig.  5oc.  All  of  the  five  members  which  are  cut,  except 
the  member  ML,  intersect  at  the  peak;  about  which  point 
the  moments  of  the  load  at  joint  7  and  the  stress  in  the  member 
ML  must  balance.  The  stress  in  the  member  NA  may  be 
determined  by  balancing  the  moments  of  all  the  forces  about  the 
joint  8  (Fig.  50^) .  After  the  stress  in  the  member  NA  has  been 
determined  and  indicated  in  Fig.  500,  the  stress  in  the  member 
KL  may  be  found  by  balancing  the  moments  of  all  the  forces 
about  the  joint  8.  When  the  stress  in  the  member  KL  has  been 
indicated  in  Fig.  5o/,  the  stress  in  the  member  JK  may  be 
determined  by  balancing  the  moments  of  all  the  forces  about  the 
joint  i.  After  the  stress  in  the  member  KL  has  been  indicated 
in  Fig.  500,  the  stress  in  LE  may  be  determined  by  balancing 
the  moments  of  all  the  forces  about  the  joint  5. 

Compute  the  stresses  in  the  members  mentioned  above  and 
see  if  they  check  with  the  stress  diagrams. 

Formulate  a  general  rule  for  applying  the  method  of  sections 
which  will  cover  all  cases  which  have  heretofore  been  considered. 

49.  The  three-hinged  arch  has  three  pin-connected  joints- 
one  at  each  of  its  two  points  of  support  and  a  third  usually 
located  at  mid-span  (Fig.  510).  The  structure,  being  free  to 
turn  at  points  A,  B  and  C,  will  obviously  collapse  even  under 
vertical  loads  unless  the  sustaining  forces  at  A  and  B  have 
horizontal  as  well  as  vertical  components.  Hence  four  unknown 
elements  for  equilibrium  must  be  determined — the  magnitude 
and  direction  of  each  reaction.  Since  only  three  independent 
equations  can  be  written  for  a  non-concurrent  system  of  forces 
acting  upon  a  rigid  body,  the  problem  of  finding  four  unknown 
elements  is  apparently  impossible.  Such,  however,  is  not  the 
case.  The  problem  is  rendered  statically  determinate  by  the 
condition  that  the  structure  is  composed  of  two  rigid  bodies, 
X  and  F,  each  free  to  turn  about  two  points. 

The  horizontal  magnitudes  of  the  external  forces  acting  on 
the  structure  as  a  whole  may  balance,  the  corresponding  vertical 
magnitudes  may  balance,  and  the  moments  about  C  of  the 
external  forces  acting  on  X  may  balance;  but  these  three  condi- 


76 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  II 


tions  do  not  certify  that  the  moments  about  C  of  the  external 
forces  acting  on  Y  are  balanced.  This  condition  provides  the 
possibility  of  making  four  independent  statements  or  equations 
concerning  the  equilibrium  of  the  structure. 

Let  us  consider  the  case  of  a  vertical  load  P  —  1,000  Ib.  at  D. 

Algebraic  Solution. — Assume  V\  and  H\  for  the  components 
of  the  reaction  at  A ;  and  F2  and  H2  for  the  components  of  the 
reaction  at  B.  Balance  the  moments  of  all  the  external  forces 
acting  on  the  structure  about  A. 


F2  = 


Similarly 


1,000  X  35 


P^IOOO 


FIG.  51  c 


The  moment  of  F2  about  C  is 

25  X  300  =  7,500  ft.-lb. 

and  the  body  F  will  rotate  counter-clockwise  about  C,  unless 
the  moment  of  F2  is  balanced  by  the  moment  of  HZ]  hence 

H2  —  — —  =  375  Ib.,  acting  to  the  left. 
20 

The  algebraic  sum  of  the  moments  of  P  and  FI  about  C  is 
25  X      700  =  17,500 
10  X  i  ,000  =  10,000 

7,500  ft.-lb.,  clockwise. 

The  body  X  will  rotate  clockwise  about  C  unless  the  moments  of 
P  and  FI  are  balanced  by  the  moment  of  HI]  hence 
7,500 


20 


=  375  Ib.,  acting  to  the  right. 


SEC.  II        APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM  77 

The  magnitude  of  HI  could  have  been  determined  by  balanc- 
ing the  horizontal  magnitudes  after  the  magnitude  of  HZ  has 
been  found.  The  magnitude  and  direction  of  each  reaction  is 
readily  determined  from  the  magnitudes  of  its  components. 

Graphic  Solution. — There  are  three  external  forces  P,  RI  and 
R2  (Fig.  5iZ>)  acting  upon  the  structure.  The  reaction  R2 
obviously  acts  through  C,  for  otherwise  the  body  Y  would  rotate 
about  C.  This  condition  determines  the  location-direction  of 
R2  which  intersects  the  force  P  at  0;  through  which  point  the 
location-direction  of  RI  must  pass  if  there  is  to  be  no  rotation 
of  the  structure  about  the  point  O.  The  magnitude-direction 
diagram  for  the  concurrent  system  at  0  is  drawn  in  Fig.  516, 
giving  the  magnitudes  of  RI  and  RI. 

In  the  case  where  both  portions  X  and  Y  are  supporting  one 
or  more  loads,  the  reactions  for  each  load  may  be  determined 
separately  after  the  manner  just  described.  The  reactions 
at  each  support  thus  found  may  then  be  combined  and  the 
resultant  reaction  for  all  loads  determined.  A  more  compli- 
cated but  somewhat  shorter  solution  is  given  in  the  next 
article. 

50.  Illustrative  Problem. — The  three-hinged  arch  (Fig.  520) 
supports  eight  loads.  Determine  the  reactions. 

Graphic  Solution. — Lay  off  the  magnitude-directions  of  the 
eight  loads  from  b  to  k  (Fig.  526).  Choose  any  convenient 
pole  o  and  imagine  that  the  nine  magnitude-directions  of  the 
components  ob  to  ok  are  drawn.  Choose  any  convenient  point 
n  on  the  location-direction  BC,  and  draw  the  location-directions 
of  the  components  OB  to  OK. 

Draw  the  magnitude-direction  bf;  and  through  the  left  and 
center  hinges  draw  lines  parallel  to  bf,  intersecting  the  location- 
directions  OB  and  OF.  Connect  these  intersections  by  the 
location-direction  OL,  which  closes  the  location-direction 
diagram  for  the  four  forces  B  to  F.  Draw  the  magnitude- 
direction  ol,  which  determines  the  magnitudes  of  two  loads  bl 
and  //;  which,  if  applied  at  the  left  and  center  hinges  respec- 
tively, will  have  the  same  effect  upon  the  equilibrium  of  the 
body  X  as  the  four  loads  B  to  F. 

The  magnitude-direction  of  the  resultant  of  the  four  loads 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  II 


F  to  K  is  fk.  Through  the  center  and  right  hinges  draw  lines 
parallel  to  fk  intersecting  the  location-directions  OF  and  OK. 
Connect  these  intersections  with  the  location-direction  OM, 
closing  the  location-direction  diagram  for  the  four  forces  F  to 
K.  Draw  the  magnitude-direction  om,  which  determines  the 
magnitudes  of  two  forces  fm  and  mk;  which,  if  applied  at  the 
center  and  right  hinges  respectively,  will  have  the  same  effect 


upon  the  equilibrium  of  the  body  Y  as  the  four  loads  F  to  K. 
For  the  sake  of  clearness  the  location-directions  BL,  LF,  FM 
and  MK  are  reproduced  in  Fig.  52^.  Two  forces  are  acting 
at  the  left  hinge — the  reaction  AB  and  the  load  BL',  the  location- 
direction  of  their  resultant  AL  must  pass  through  the  left  and 
center  hinges.  Therefore  through  /  draw  the  magnitude- 
direction  parallel  to  the  location-direction  AL.  Similarly 
the  location-direction  of  the  resultant  of  the  load  MK  and  the 
reaction  KA  acting  at  the  right  hinge  is  M A ,  which  must  pass 
through  the  center  and  right  hinges.  Hence  through  m 


SEC.  II        APPLICATION    OF    PRINCIPLES    OF    EQUILIBRIUM 


79 


draw  the  magnitude-direction  parallel  to  the  location-direction 
MA  intersecting  the  magnitude-direction  through  /  at  a. 

The  location  of  a  fixes  the  magnitude-directions  ka  and  ab 
which,  when  drawn,  will  close  the  magnitude-direction  diagram 
for  the  external  forces  and  completely  determine  the  reactions. 

Draw  the  stress  diagram  and  note  that  the  diagram  gives  a 
check  on  itself  when  the  center  hinge  is  reached. 

It  may  be  well  to  note  that  the  load  FG  acting  at  the  center 
hinge,  which  was  combined  with  the  loads  on  the  body  F, 
might  have  been  combined  with  the  loads  on  the  body  X 
without  affecting  the  final  results. 

Prove  the  statement  in  the  preceding  paragraph  by  a  solution. 

Algebraic  Solution. — Replace  each  inclined  load  by  its  H-  and 
F-components.  Assume  V\  and  HI  for  the  components  of  the  left 
reaction;  and  Vz  and  HZ  for  the  components  of  the  right  reaction. 
Balance  the  moments  of  all  the  forces  acting  on  the  structure 
about  the  left  support. 


35  X 

[,2OO  = 

42,000 

65  X 

720  = 

46,800 

75  X 

500  = 

37,500 

30  X 

[,350  = 

4O,5OO 

60  X 

,200  = 

72,000 

9oX 

,200  = 

IO8,OOO 

120  X 

,510  = 

l8l,200 

150  X 

,300  = 

195,000 

180  X 

,6OO  = 

288,000 

210  X 

QOO  = 

l89,OOO 

240)1 

,2OO,OOO 

5,000  =  V* 
Balance  the  moments  about  the  right  support, 


30  X 

900  = 

27,000 

60  X 

,600  = 

96,OOO 

90  X 

,300  = 

117,000 

120  X 

,510  = 

l8l,2OO 

150  X 

,200  = 

l8o,OOO 

180  X 

,2OO  = 

2l6,OOO 

210  X 

,350  = 

283,500 

240  X 

900  = 

216,000 

I 

,316,700 

I26,3OO 

240)l 

,I9O,4OO 

4,960 

75  X     500  =    37,500 

65  X     720  =    46,800 

35  X  1,200  =    42,000 

126,300 


Vi 


8o 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  II 


Balance  the  moments  of  the  forces  on  the  body  Y  about  the  center  hinge, 
30  X  1,300  =    39,000  120  X  5>ooo  =  600,000 

60  X  1,600  =    96,000  216,000 

90  X     900  =    81,000  80)384,000 


216,000 


4,800 
acting  to  the  left. 


80)190,400 


Balance  the  moments  of  the  forces  on  the  body  X  about  the  center  hinge, 
30  X  1,200  =    36,000  120  X  4,960  =  595,200 

60  X  1,200  =    72,000  404,800 

5  X      500  =      2,500 
90  X  1,350  =  121,500 

15  X     720=    10,800  acting  to  the  right. 

1  20  X     900  =  108,000 

45  X  1,200  =    54,000 

404,800 

Check  by  balancing  the  horizontal  magnitudes, 
2,380 

1,200 

720 
500 

4,800  =  4,800 
Compare  these  results  with  the  H-  and  F  -components  of  ka  and  db  (Fig.  526). 

51.  The   cantilever  bridge   supporting   vertical  loads   and 
resting  upon  four  vertical  supports  (Fig.  53  a)  appears  to  be 


statically  indeterminate,  for  there  are  four  magnitudes  to 
determine — one  at  each  support;  but  the  indeterminateness 
disappears  when  we  consider  that  the  structure  is  composed  of 
three  rigid  bodies  AC,  CD  and  DF  fastened. together  by  pin- 
connected  joints  at  C  and  D.  The  portions  AC  and  DF  are 


SEC.  II        APPLICATION    OF    PRINCIPLES    OF    EQUILIBRIUM 


8l 


called  cantilever  trusses;  AB  and  EF  are  the  anchor  arms  and 
BC  and  DE  are  the  cantilever  arms.  The  portion  CD  is 
called  the  center  or  suspended  truss.  The  interaction  of  one 
part  of  the  structure  upon  another  is  seen  more  clearly  by  a 
consideration  of  Fig.  536,  where  the  center  truss  is  shown  as  if  it 
were  suspended  from  the  cantilever  arms. 

Loads  applied  at  points  i ,  2  and  3  are  supported  at  A  and  B ; 
and  that  part  of  the  structure  represented  by  AGHB  performs 
the  function  of  a  simple  truss.  Loads  at  points  4,  5  and  6  are 
also  supported  at  A  and  B\  but  in  this  case  the  reaction  at  A 
is  negative,  i.e.,  acts  downward.  Consequently,  the  structure 
must  be  provided  with  anchors  into  the  masonry  at  A  and  F. 

Suppose  that  a  load  of  1,000  Ib.  is  placed  at  point  7.  There 
is  a  tension  of  750  Ib.  in  CC'  causing  a  negative  reaction  of  750 
Ib.  at  A,  and  an  upward  pressure  of  1,500  Ib.  at  B.  Similarly 
there  is  a  negative  reaction  of  250  Ib.  at  F,  and  an  upward 
pressure  of  500  Ib.  at  E.  The  reactions  at  E  and  F  for  loads 
between  D  and  F  are  determined  in  the  same  manner  as  for  the 
left  cantilever  truss. 

The  foregoing  analysis  as  applied  to  the  structure  in  Fig.  536 
is  equally  applicable  in  Fig.  530. 

52.  Inclined  Loads. — In  the  next  chapter  we  shall  have 
occasion  to  consider  the  effect  of  loads  acting  normal  to  the 


8  ft  6'=  48' >J 


slope  of  the  roof,  the  roof  truss  being  supported  on  walls,  as 
illustrated  in  Fig.  540.  The  resultant  of  the  five  loads  is  5,200 
Ib.  Its  H-  and  F-components  are  2,000  Ib.  and  4,800  Ib. 
respectively. 

Let  HI  and  V\  represent  the  H-  and  F-components  at  A,  and 
H-2.  and  Vz  represent  the  corresponding  components  at  B.     Since 


82  THEORY   OF  FRAMED   STRUCTURES  CHAP.  II 

the  points  of  support  are  level  and  the  //-components  act  in  a 
straight  line,  V\  and  F2  may  be  determined  independently  of 
Hi  and  Z72. 

Balance  the  moments  of  all  the  forces  about  the  point  A, 
5,200  X  13 


48 

4,800 
1,408 


=  1,408  =  V. 


3,392  =  Vi 

Check  the  value  of  V\  by  balancing  the  moments  of  all  the 
forces  about  B. 

The  magnitudes  of  HI  and  #2  acting  in  a  straight  line  cannot 
be  determined  (as  in  the  case  of  a  three-hinged  arch  consisting 
of  two  rigid  bodies)  by  the  principles  of  statics.  Their  sum  is 
2,000  lb.,  and  the  magnitude  of  each  depends  somewhat  upon 
the  manner  in  which  the  truss  is  supported.  If  the  truss  has  a 
roller  support  at  B  and  we  assume  no  friction,  then 

HI  —  2,000 
and  H2  =         o 

If  the  truss  has  a  roller  support  at  A  and  we  assume  no  friction, 

then  HI  =         o 

and  HZ  =  2,000 

If  we  assume  that  the  horizontal  thurst  is  resisted  equally  at 

A  and  B,  then 

HI  =  1,000 

and  HI  =  1,000 

If  we  assume  that  the  horizontal  component  of  each  reaction  is 
proportional  to  its  vertical  component,  then 

-       *i  -5^X2,000-1,413" 


and        ,, 

In  the  latter  case  the  resultant  of  the  H-  and  F-components  at 

each  support  has  a  direction  parallel  to  the  resultant  of  the 

loads. 

In  Fig.  546  the  load  line  cd  is  laid  off  and  the  magnitude- 


SEC.  Ill       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM  83 

direction  diagram  lor  the  external  forces  is  closed  from  d  to  c 
in  four  different  ways;  corresponding  to  the  various  assump- 
tions which  may  be  made  regarding  the  magnitudes  of  H\  and 
H^.  Since  the  vertical  magnitudes  are  constant  for  any  as- 
sumed magnitudes  for  HI  and  #2,  the  points  e\,  £2,  es  and  e± 
fall  on  a  horizontal  line. 

Complete  the  stress  diagram.  How  do  the  different  as- 
sumptions affect  the  stresses? 

SEC.  III.     BEAMS:  SHEAR  AND  BENDING  MOMENT  DIAGRAMS 

53.  A  beam  is  any  part  of  a  structure  or  the  structure  itself 
as  a  whole,  which  resists  the  action  of  transverse  forces  and  is 
bent  by  them.     The  forces  may  act  at  right  angles  to  the  axis 
of  the  beam,  or  they  may  be  inclined  at  any  angle.     When  the 
forces  are  inclined  they  may  be  resolved  at  the  axis  of  the  beam 
into  rectangular  components  normal  and  parallel  to  the  axis. 
The  components  parallel,  i.e.,  in  line  with  the  axis  produce 
direct  tension  or  compression  in  the  beam ;  while  the  normal  or 
transverse  components  produce  the  beam  action  or  bending. 
The  effect  of  the  transverse  forces  is  treated  under  the  headings 
of  shear  and  bending  moment. 

54.  The  shear  at  any  normal  section  of  a  beam  is  the  alge- 
braic sum  of  all  the  transverse  forces  acting  on  one  side  (either 
side)  of  the  section.     Since  the  algebraic  sum  of  all  the  trans- 
verse forces  acting  upon  a  beam  in  equilibrium  is  zero  (S7  = 
o);  it  is  evident  that  the  shears  on  the  two  sides  of  a  section 
have  equal  magnitudes  but  are  opposite  in  sense.     The  shear 
is  called  positive,  when  acting  upward  on  the  left  (of  the  section) 
or  downward  on  the  right;  negative,  when  acting  downward  on 
the  left  or  upward  on  the  right. 

55*  The  bending  moment  at  any  normal  section  of  a  beam  is 
the  algebraic  sum  of  the  moments  of  all  the  forces  acting  on  one 
side  (either  side)  of  the  section;  taken  about  the  center  of 
gravity  of  the  section  as  an  axis.  Since  the  algebraic  sum  of  the 
moments  about  any  point  of  all  the  forces  acting  upon  a  beam 
in  equilibrium  is  zero  (I.M  =  o) ;  it  is  evident  that  the  bending 
moments  on  the  two  sides  of  a  section  have  equal  magnitudes 


84  THEORY    OF   FRAMED    STRUCTURES  CHAP.  II 

but  are  opposite  in  rotation — the  one  being  clockwise  and  the 
other  counter-clockwise.  The  bending  moment  is  called 
positive  when  clockwise  on  the  left  of  and  about  the  section,  or 
counter-clockwise  on  the  right  of  and  about  the  section;  nega- 
tive, when  counter-clockwise  on  the  left  of  and  about  the  sec- 
tion, or  clockwise  on  the  right  of  and  about  the  section. 

56.  The  Bending  Moment  Determined  from  the  Location- 
direction  Diagram. — In  general  the  shear  and  bending  moment 
vary  from  section  to  section  along  a  beam.  They  may  be 
represented  conveniently  for  inspection  by  diagrams,  the  ordi- 
nates  of  which  represent  the  shear  or  bending  moment  at 
various  sections  of  the  beam.  Each  diagram  has  a  base  or 
datum  line.  Positive  values  are  laid  off  above,  and  negative 
values  below  this  line.  These  diagrams  have  a  definite  rela- 
tion to  the  magnitude-direction  and  location-direction  dia- 
grams, which  are  drawn  in  the  graphic  solution  of  a  system  of 
parallel  forces.  The  diagrams  in  Figs.  220,  and  226  were  drawn 
to  determine  the  magnitude  of  Q  and  the  magnitude  and  direc- 
tion of  P.  These  diagrams  are  reproduced  with  some  addi- 
tions in  Fig.  55.  The  shear  at  any  section  may  be  found  from 
the  magnitude-direction  diagram.  The  shear  on  any  section 
between  the  left  reaction  and  the  first  load  is  fa  =  +5  Ib. 
The  shear  on  any  section  between  the  first  and  second  loads  is 
ft  =  -j-^  Ib.  The  shear  on  any  section  between  the  second 
and  third  loads  is/c  =  —2  Ib.,  etc. 

The  bending  moment  at  any  section  x  between  the  first  and 
second  loads  is 

Mx  =  pP  -  sS 

Produce  the  location-directions  ob  and  of,  to  intersect  at  /. 
The  point  7  locates  the  resultant  of  the  two  forces  P  and  5; 
and  the  magnitude  of  this  resultant  is  the  magnitude-direction 
fb,  therefore 

pp  -  SS  =  fb.r 
The  triangles  GIJ  and  bof  are  similar,  hence 

y:r  ::fb  :  h 

then  fb.r  =  yh 

or  M  x  =  yh 


SEC.  Ill       APPLICATION    OF   PRINCIPLES    OF    EQUILIBRIUM 


Fl&.S&a 


86  THEORY   OF   FRAMED    STRUCTURES  CHAP.  II 

Hence,  the  bending  moment  at  any  section  equals  the  correspond- 
ing ordinate  y  in  the  location-direction  diagram,  multiplied  by 
the  horizontal  component  of  the  forces  in  the  magnitude-direc- 
tion diagram.  The  ordinates  y  are  measured  in  feet,  and  the 
component  h  is  measured  in  pounds;  hence,  the  bending  moment 
is  expressed  in  foot-pounds. 

If  the  reactions  had  been  determined  algebraically,  the 
magnitude-direction  diagram  could  have  been  drawn  and  closed; 
locating  the  point  /,  before  the  location-direction  diagram  was 
drawn.  It  would  then  have  been  possible  to  have  chosen  a 
point  on  a  horizontal  line  through/  for  the  pole  o,  (Fig.  560). 
In  this  case  the  location-direction  of  becomes  a  horizontal  line 
as  shown  in  Fig.  566;  and  the  bending  moment  at  any  section 
equals  10  Ib.  times  the  number  of  feet  scaled  on  the  correspond- 
ing ordinate  in  the  location-direction  diagram.  It  has  been 
shown  that  the  shear  at  any  section  may  be  obtained 
from  the  magnitude-direction  diagram.  The  shear  at  various 
sections  may  be  represented  more  clearly  by  the  shear  diagram 
(Fig.  $6c).  The  datum  or  base  line  is//",  and  the  shear  at  any 
section  is  represented  by  the  ordinate  at  that  section. 

It  is  now  clearly  seen  that  the  slope  or  bevel  of  each  line 
in  the  location-direction  diagram  is  proportional  to  the  shear;  for 
if  h  =  10  Ib.  is  taken  as  the  horizontal  side  of  each  bevel,  the 
vertical  side  of  the  bevel  will  represent  the  shear.  Having 
established  the  principle  that  the  bending  moment  may  be 
derived  from  the  location-direction  diagram,  and  that  the  slope 
of  any  side  of  the  location-direction  diagram  is  a  function  of 
the  shear;  we  shall  proceed  to  consider  the  topic  of  shear  and 
moment  diagrams  from  another  point  of  view. 

57.  Shear  and  Bending  Moment  Diagrams.— The  graphic 
method  outlined  in  the  previous  article,  whereby  the  shear  and 
bending  moment  may  be  obtained  from  the  magnitude-direc- 
tion and  location-direction  diagrams,  is  not  the  one  most  com- 
monly used  by  practicing  engineers.  This  is  especially  true 
when  the  beam  under  consideration  supports  a  uniform  load 
over  the  whole  or  a  portion  of  its  length.  A  more  convenient 
method,  which  is  semi-graphical,  will  now  be  presented. 


SEC.  Ill      APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM  87 

58.  Illustrative  Problems. 

i .  Draw  the  shear  and  bending  moment  diagrams  for  the  beam 
shown  in  Fig.  57. 

The  shear  diagram  is  drawn  first  because  it  is  the  simpler  of 
the  two.  Draw  a  vertical  line  through  the  location  of  each 
load  and  reaction,  and  let  ST  be  the  base  line  for  the  shearing 
forces.  The  shear  at  any  section  between  the  left  reaction  and 
the  first  load  is  +5  Ib.  Lay  off  SA  =  +5  Ib.  to  any  convenient 
scale  and  draw  AB  parallel  to  ST.  On  the  ordinate  through  B, 
lay  off  downward  BC=  —2  Ib.  and  draw  CD.  Lay  off  DF  = 


—  5  Ib.  and  draw  FG.  Continue  in  like  manner  to  H  on  the 
ordinate  through  the  right  support,  and  note  that  HT  scales 
9  Ib.,  which  equals  the  right  reaction.  The  broken  line  SABC 
...  HT  is  the  shear  diagram,  since  the  measure  of  any 
ordinate  gives  the  shear  at  the  corresponding  section  of  the 
beam. 

For  practical  purposes  it  is  seldom  necessary  to  draw  the 
shear  diagram  to  scale.  A  sketch  will  usually  serve  equally  as 
well. 

We  now  proceed  to  sketch  the  bending  moment  diagram  on 
the  base  line  UV.  The  bending  moment,  in  accordance  with 


88  THEORY   OF   FRAMED    STRUCTURES  CHAP.  II 

its  definition,  is  zero  at  each  end  of  the  beam;  consequently  the 
diagram  may  be  said  to  begin  at  U  and  end  at  V.  In  Article 
56  it  was  shown  that  the  slope  of  the  bending  moment  diagram 
was  proportional  to  the  shear.  The  shear  from  A  to  B  is  posi- 
tive, hence  the  line  UJ  is  sketched  having  a  positive  slope  (i.e., 
upward  to  the  right) .  The  shear  from  C  to  D  is  also  positive, 
but  not  as  great  as  the  shear  from  A  to  B ;  hence  JK  is  sketched 
having  a  positive  but  lesser  slope  than  the  line  UJ.  The  shear 
from  F  to  G  is  negative  and  the  line  KL  is  sketched  with  a 
negative  slope  (i.e.j  downward  to  the  right).  The  slopes  of 
LN  and  NV  are  negative,  NV  having  the  steeper  slope.  Thus 
a  general  idea  of  the  variation  in  bending  moment  may  be  ob- 
tained from  a  simple  sketch. 

The  horizontal  unit  of  measure  in  both  diagrams  is  i  ft.  The 
vertical  unit  of  measure  is  i  Ib.  in  the  shear  diagram,  and  i  ft.- 
Ib.  in  the  bending  moment  diagram.  Let  i  ft.  be  taken  as  the 
length  of  the  horizontal  side  of  the  triangle,  representing  the 
slope  or  bevel  of  each  line  in  the  bending  moment  diagram ;  then 
the  length  of  the  vertical  side  of  each  triangle  will  be  measured 
in  foot-pounds,  and  the  ratio  of  the  vertical  side  to  the  horizontal 
side  will  represent  pounds  and  equal  the  shear.  Thus  the 

slope  of  the  line  UJ  is 77 — '  =  5  Ib.;  the  slope  of  the  line 

KL    is  -2    *'"    '  =  -2  Ib.,  etc.     The  length  of  the  critical 

ordinates  JQ,  KR,  etc.,  may  be  obtained  from  the  similarity  of 
triangles. 

UQ:i'::QJ:s'# 
or  6':i'i:QJ:S'# 

whence  QJ  =  -  -^7^  =  6'  X  5#  =  3<>'#  =  area  SABP 

From  this  it  is  clear  that  the  difference  in  lengths  of  any  two 
ordinates  in  the  bending  moment  diagram  equals  the  area  of  the 
shear  diagram  between  the  ordinates.  The  lengths  of  the 
consecutive  ordinates  may  be  computed  as  follows: 


SEC.  Ill       APPLICATION    OF   PRINCIPLES    OF    EQUILIBRIUM 


89 


+  5X6 


o  =  ordinate  at  U 
+30 


+30  =  JQ 

+3X5  =  +15 

+45  = 
-2X4=  -8 

+37  = 
—  5X2=  —IP 

+  27  =  NX 
-9X3=  -27 

o  =  ordinate  at  V. 

This  method  of  calculation  checks  itself,  and  is  particularly 
helpful    when  the  shear  and  bending  moment  diagrams  are 


FIG.  58. 

simply  sketched  and  not  drawn  to  scale.  After  the  critical 
ordinates  have  been  computed,  the  bending  moment  diagrams 
may  be  drawn  to  scale. 

In  going  from  left  to  right  along  the  beam,  it  is  obvious 
that  the  bending  moment  is  increasing  as  long  as  the  shear  is 
positive  and  the  lines  in  the  bending  moment  diagram  continue 
to  have  a  positive  slope.  At  the  section  where  the  shear  changes 
from  positive  to  negative,  the  slope  changes  from  positive  to 
negative  and  the  bending  moment  is  a  maximum. 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  II 


2.  The  shear  and  bending  moment  diagrams  for  a  cantilever 
beam  are  sketched  in  Fig.  58.  Between  A  and  B  the  shear  is 
negative  and  GH  has  a  negative  slope.  At  B  the  shear  changes 
from  negative  to  positive,  hence  HK  has  a  positive  slope,  and  a 
maximum  negative  bending  moment  occurs  at  B.  At  D  the 
shear  changes  from  positive  to  negative,  giving  a  maximum 
positive  bending  moment.  The  computations  for  the  bending 
moments  at  critical  ordinates  follow : 

o  at  A 
—  30  X  10  =  —300 

—  300  at  B 
+  132  X  6  =  +792 

+492  at  C 
+  72  X  10  =  +720 

+  1,212  at  D 
-48  X    8  =  -384 

+828  at  E 
-136  X  6  =  -828 


POO  Ik  per  ft 


U 


3.  The  beam  in  Fig.  59  supports  a  uniform  load  of  200  Ib. 


SEC.  Ill      APPLICATION   OF   PRINCIPLES    OF   EQUILIBRIUM 


per  foot.  The  shear  diagram  starts  with  a  positive  ordinate  = 
i, 600  Ib.  at  S,  decreasing  uniformly  200  Ib.  in  each  foot  from  S 
to  C.  Beyond  C  the  ordinates  are  negative,  and  increasing 
200  Ib.  per  foot  to  — 1,600  Ib.  at  T.  Since  the  ordinates  in  the 
shear  diagram  are  uniformly  varying,  the  slope  of  the  moment 
diagram  will  be  uniformly  varying;  and  the  bending  moment 
diagram  will  be  represented  by  a  curved  line  instead  of  a  series 
of  broken  lines.  The  ordinates  in  the  shear  diagram  are  positive 


1     h 

if   \ 

A i. 


FIG.  6o&. 

over  the  left-half  of  the  beam,  decreasing  to  zero  at  the  center; 
hence  the  bending  moment  curve  has  a  positive  slope  at  U, 
which  decreases  to  zero  at  K,  where  the  tangent  to  the  curve  is 
horizontal.  Beyond  K  the  curve  has  an  increasing  negative 
slope. 

Any    ordinate    in  the  bending  moment  diagram  is  easily 
determined  from  the  area  of  the  shear  diagram;  thus 

//  =  area  SABF  =  4,800  ft.-lb. 
KH  =  area  SAC     =  6,400  ft.-lb. 

The  bending-moment  curve  is  a  parabola,  as  will  be  shown  in  the 
following  article. 

59.  The  Structural  Engineer's  Parabola. — The  mathemati- 
cian usually  derives  the  equation  of  the  parabola  with  the 


92  THEORY    OF   FRAMED    STRUCTURES  CHAP.   II 

origin  at  the  vertex  and  the  axis  of  symmetry  horizontal,  as 
shown  in  Fig.  6oa.  Under  these  conditions  the  general  equation 
of  the  parabola  is 

F2  =  kX  (i) 

where  X  and  Y  are  variables  and  k  is  a  constant. 

The  structural  engineer  finds  it  more  convenient  to  choose 
some  point  not  at  the  vertex  for  the  origin,  with  the  axis  of 
symmetry  vertical  as  shown  in  Fig.  606.  Let  P  be  any  point 
on  the  curve  in  either  Fig.  6oa  or  b.  Let  X  and  Y  represent  the 
coordinates  of  P  when  the  origin  is  at  the  vertex  M,  and  let  x 
and  y  represent  the  coordinates  when  the  origin  is  at  some  other 
point  E.  When  the  point  P  is  at  N 

X  =  h,  and  Y  =  ~ 
Substituting  these  values  in  Eq.  (i) 

?-** 

4 

°r  *  =  i* 

Hence  the  mathematician's  equation  of  this  particular  parabola 
is 


Substituting  X=  h  —  y,  and  F  =  x 


inEq.(2)  , 

or  y  =       x  (I  -  x)  (3) 

Equation  (3)  may  be  called  the  engineer's  equation  of  the 
parabola,  in  the  same  sense  that  Eq.  (2)  may  be  said  to  represent 
the  mathematician's  equation  of  the  same  curve.  Equation  (3) 
may  be  written  in  the  more  general  form 

y  =  cx(l  -  x)  (4) 

where  x  and  y  are  variables  and  c  is  a  constant.  Equation  (4) 
shows  that  one  variable  (y)  is  proportional  to  the  product  of  two 


SEC.  Ill       APPLICATION    OF   PRINCIPLES    OF    EQUILIBRIUM  93 

other  variables  (x)  and  (I  —  x)  whose  sum  (/)  is  constant. 
Any  equation  which  can  be  expressed  in  this  form  is  a  parabolic 
equation. 

The  beam  in  Fig.  61  supports  a  uniform  load  of  w  pounds  per 
unit  length.  The  bending  moment  at  any  distance  x  from  the 
left  support  is 

wlx       wx2       w     f  ,  v 

Mx  =  —       —    =  -  *  (/  -  x)  (5) 

This  is  a  parabolic  equation  and         i   w* per  lin.fi: 

1 1      i        i  •  T  wl  riiiiiiiiii  INI  HIM  ii  illinium  INI  mi  urn 

the  bending  moment   diagram     j.lT^ 

is  a  parabola.     The   following  " "  l~x  ' 

important  rule  may  be  deduced         U--  -  i 

from  Eq.  (5). 

The  bending  moment  at  any 
section  of  a  uniformly  loaded 
beam  equals  one-half  the  load 

FIG.  61. 

per   unit  of  length,  times   the 

product  of  the  two  segments  into  which  the  section  divides  the 

span. 

Thus  the  bending  moment  at  //  (Fig.  59)  is 

100  X  4  X  12  =  4,800 

Two  Methods  of  Constructing  a  Parabola. — There  are 
several  methods  by  which  a  parabola  may  be  constructed,  but 
only  two  will  be  considered  here:  (a)  an  algebraic  method  by 
points;  and  (b)  a  graphic  method  by  tangents. 

(a)  Algebraic  Method. — Equation  (3)  may  be  written  in  the 
form 


from  which  the  following  observation  may  be  made  in  connec- 
tion with  Fig.  59.  Any  two  ordinates  //  and  KH  are  to  each 
other  as  the  products  of  the  two  parts  into  which  each  ordinate 
divides  the  base  line  UV.  This  equation  gives  a  clue  to  a  very 
simple  method  of  locating  any  desired  number  of  points 
through  which  a  parabola  is  to  be  drawn.  Suppose  it  is 
desired  to  locate  seven  points  on  the  curve.  Divide  the  line 
UV  into  eight  equal  parts  by  seven  points.  The  ordinate  at 


94 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  II 


each  point  is  proportional  to  the  product  of  the  number  of 
parts  on  either  side  of  it,  thus: 


1X7  = 
2X6  = 


7 

12 

i5 


3X  5= 
4X4=  16 

5X3=  i5 
6X2=12 

7X1=7 

Now  if  we  wish  the  middle  ordinate  to  equal  6,400  instead  of  16, 
we  multiply  that  ordinate  and  all  the  others  by  400,  whence 

7  X  400  =  2,800 
12  X  400  =  4,800 

15  X  400  =  6,000 

16  X  400  =  6,400,  etc. 

(b)  Graphic  Method. — If  the  tangents  at  the  extremities  of  a 
parabolic  curve  are  known,  the  parabola  may  be  constructed 


FIG.  62. 

as  shown  in  Fig.  62.  Suppose  we  have  two  tangents  AC  and 
EC  and  the  points  of  tangency  A  and  B.  Divide  AC  into  any 
number  of  equal  parts,  and  BC  into  the  same  number  of  equal 
parts.  Number  the  points  from  A  to  C  and  from  C  to  B. 
Connect  i-i,  2-2,  etc.  The  parabolic  curve  lies  tangent  to 
these  lines  as  shown. 


SEC.  Ill      APPLICATION   OF   PRINCIPLES    OF   EQUILIBRIUM  95 

Draw  tangents  through  U  and  V  (Fig.  59)  intersecting  at  D. 
It  may  be  observed  from  the  shear  diagram  that  the  positive 
slope  of  the  tangent  UD  is  1,600  ft.-lb.  vertical  to  i  ft.  hori- 
zontal. The  tangent  DV  has  a  corresponding  negative  slope, 
hence  the  intersection  D  is  directly  above  K  and  the  ordinate 
HD  =  12,800  ft.-lb.  Construct  the  parabolic  bending  moment 
diagram  on  these  tangents,  and  check  the  lengths  of  several 
ordinates  in  the  moment  diagram  by  the  algebraic  method. 

60.  Illustrative  Problem. 

The  shear  and  bending  moment  diagrams  for  a  beam  (Fig.  63) 
supporting  a  uniform  load  over  a  portion  of  its  length,  have 
several  interesting  properties.  The  shear  diagram  is  drawn  first, 
and  the  section  at  which  the  shear  changes  from  positive  to 
negative  is  located.  The  ordinates  in  the  shear  diagram 
indicate  that  the  moment  diagram  is  composed  of  the  straight 
line  OA,  the  parabolic  curve  A  BCD,  and  the  straight  lines  DE 
and  EX.  The  maximum  ordinate  RC  is  at  the  section  of  zero 
shear.  The  following  ordinates  are  obtained  from  the  area  of 
the  shear  diagram : 

o  at  0 
+  4,000  X  5  =  +20,000 

+  20,000  =  NA 
|(4,ooo  +  1,000)15  =  +37>5°° 

+  57,500  =  JB 
+  1,000  X  \  X  5  =     +2,500 

+  60,000  =  RC 
—  2,000  X  J  X  10  =  —10.000 

+  50,000  =  WD 
—  2,000  X  15  =  —30,000 

+  20,000  =  ZE 
—  5,000  X  4  =  —20,000 

o  at  X 

Lay  off  to  scale  the  ordinates  NA,  WD  and  ZE\  and  draw  the 
lines  OA,  DE  and  EX.  The  line  OA,  having  the  same  slope 
at  A  as  the  parabola,  is  tangent  to  it ;  likewise  the  line  DE  is 
tangent  to  the  parabola  at  D.  If  the  lines  OA  and  ED  are 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.   II 


produced,  they  will  intersect  at  F ;  a  point  directly  under  the 
center  of  the  uniform  load.     If  the  uniform  load  of  6,000  Ib. 


4-000 

K— - 


4000 


•« -X    > 


A^t 


If 

M- 


Z 


Ibs.per  foot 


30001k 


/y/WvZv///^^ 


S4'-Off- 


>> 

/^* 


-2000 


5000 


-5000 


3000 


B, 


FIG.  63. 


were  concentrated  at  its  center,  the  bending  moment  diagram 
would  be  OF  EX.  The  parabola  may  now  be  drawn  on  the 
tangents  FA  and  FD. 


SEC.  Ill       APPLICATION    OF   PRINCIPLES    OF    EQUILIBRIUM  97 

The  parabolic  segment  ABDG  has  several  interesting  prop- 
erties. Let  PS  represent  any  ordinate  in  the  bending  moment 
diagram  between  AN  and  DW,  and  let  x  represent  the  distance 
of  this  ordinate  from  the  left  support;  then  the  bending 
moment  at  x  is 

M.  =  PS  =  4,000*  - 


5,ooo#  —  2,500  (i) 

Let  QS  =  y  be  the  ordinate  to  the  line  AD,  then  from  similar 
triangles 

QT:DH::TA:HA 

QT  —  y  —  AN  =  y  —  20,000 

DH  =  DW  -  AN  =  30,000 

TA  =  x  -  5 

HA  =  30 

therefore  y  —  20,000:30,000::^  —  5:30 

or  y  —  i,ooox  +  15,000 

then    PQ  =  Mx  —  y  —  m  =  —  ioox2  -\-  4,000*  —  17,500 
Let     AT  =  x  —  $  =  v 
or  x  =  v  +  5 

then      m  =  -100(0  +  5)2  +  4,000(2;  +  5)  -  17,500 
or          m  =  100^(30  —  v)  (2) 

Equation  (2)  is  also  the  expression  for  the  bending  moment  m 
at  any  section  of  a  beam  30  ft.  long,  supporting  a  uniform  load 
of  200  Ib.  per  foot  when  the  section  is  a  distance  v  from  either 
support.  Therefore  the  lengths  of  corresponding  ordinates 
in  the  two  parabolic  segments  ABDG  and  AiBiDid  are  equal. 
The  areas  of  the  two  segments  are  equal  and  the  centers  of 
gravity  are  similarly  situated. 

DH 

The  slope  of  the  line  AD  is      -j  =  1,000  Ib.     The  shear 


ordinate  KL  is  1,000  Ib.;  hence  the  line  UV,  which  is  tangent  to 
the  parabola  at  B,  has  a  slope  of  1,000  Ib.,  and  is  therefore 
parallel  to  the  line  AD.    The  point  B  bisects  the  line  FG. 
61.  Problems. 

1-7.  Determine  the  reactions  algebraically,  sketch  the  shear  and  bending 
moment  diagrams,  compute  the  values  for  critical  ordinates  and  draw  the 
diagrams  to  scale  for  the  beams  shown  in  Figs.  64  to  70  inclusive.  The  loads  A 


98 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  II 


and  B  in  Fig.  65  are  equal.  In  Fig.  66,  A,  B  and  C  are  frictionless  pegs.  The 
rope  passes  over  the  upper  pegs  and  under  the  lower  one.  The  beams  in  Figs. 
67,  68  and  69,  are  fixed  in  a  wall  at  the  left  end. 


1800* 


as 

'  5>       J^  $'  j 

FIG.  64. 

t  I         f     f     \seoo* 

uv»U— -•£'  — j<*$£»fefyf4tu..  /0'-- 


FIG.  65. 


80* 


FIG.  66. 


1 


FIG.67 

300  Ib. per  ft. 


W 

FIG. 68 


800  Ik  per  ft 


10' 

F1G.69. 


8.  A  timber  20  ft.  long,  weighing  40  Ib.  per  linear  foot  is  floating  in  water, 
and  supports  200  Ib.  4  ft.  from  the  left  end  and  150  Ib.  2  ft.  from  the  right  end. 
Draw  the  shear  and  bending-moment  diagrams. 


SEC.  IV       APPLICATION    OF    PRINCIPLES    OF    EQUILIBRIUM  99 

9.  A  timber  80  ft.  long,  floating  in  water,  supports  three  loads  of  80  Ib.  each, 
one  at  the  center  and  one  at  each  quarter  point.  Neglect  the  weight  of  the 
timber  and  draw  shear  and  bending  moment  diagrams. 


10.  A  beam  30  ft.  long  is  supported  at  each  end  A  and  C.  A  uniform  load 
of  100  Ib.  per  foot  extends  from  A  to  B,  and  a  uniform  load  of  225  Ib.  per  foot 
extends  from  B  to  C.  The  bending  moment  is  a  maximum  at  B.  Locate  B, 
and  draw  the  shear  and  bending-moment  diagrams. 

SEC.  IV.     FRAMES  HAVING  MEMBERS  WHICH  PERFORM  THE 
FUNCTIONS  OF  A  BEAM 

62.  The  framed  structures  heretofore  discussed  have  sup- 
ported loads  applied  only  at  the  joints.     This  is  one  of  the  four 
conditions  mentioned  in  Article  40,  which  must  be  met  if  the 
members  of  a  structure  are  subject  to  longitudinal  stresses 
(tension  or  compression)  only.     In  order  to  prepare  the  way 
for   an  analysis  of  a  structure  composed  of  a  roof  truss  and 
columns,  in  which  the  columns  also  serve  as  beams  when  the 
structure  is  resisting    inclined  loads;  it  will  be  advisable  to 
consider  several  types  of  simple  frames  which  resist  the  action 
of  loads  applied  not  only  at  the  joints,  but  also  at  intermediate 
points.     Such  members  must  be  designed  to  resist  shearing 
and  bending  stresses,   as  well  as  longitudinal  stresses.     The 
analysis  of  a  structure  of  this  type  may  be  made  by  a  process  of 
dissection,   whereby  each  member  is  sketched  separately  and 
the  forces  acting  thereon  are  shown. 

63.  Illustrative  Problem. 

The  frame  shown  in  Fig.  71,  hinged  at  A  and  resting  on  a 
roller  at  B,  supports  a  load  of  300  Ib.  at  E.  We  find  first  the 
reactions  or  the  forces  necessary  for  the  equilibrium  of  the 
frame  as  a  whole.  On  account  of  the  roller  at  B}  the  reaction 
there  is  horizontal;  consequently  there  is  a  vertical  force  of  300 
Ib.  acting  upward  at  A,  and  a  horizontal  force  equal  and  opposite 
to  the  horizontal  force  at  B.  Balance  the  moments  about  A. 

300  X  6  = 
24         :  75 


IOO 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  II 


The  horizontal  force  of  75  Ib.  acts  to  the  left  at  B,  and  to  the 
right  at  A .     Now  sketch  each  member  separately. 

The  vertical  member  resists  the  action  of  forces  at  four  points, 
A,  B,  C  and  D.  The  forces  at  A  and  B  are  known.  The  un- 
known forces  at  C  and  D  are  the  result  of  the  interaction  of  the 
members,  and  are  represented  by  horizontal  and  vertical  com- 
ponents. The  horizontal  member  resists  the  action  of  forces 
at  three  points,  the  force  at  E  being  known.  Indicate  hori- 
zontal and  vertical  components  at  D'  and  F.  The  inclined 
member  resists  forces  at  two  points  Ff  and  C",  as  shown.  Bal- 


* — ^ 


-UL75 


n.75 

B  •*— 


300 


F        D' 


300 


w 


FIG.  71. 


ance  the  moments  of  the  forces  acting  on  the  horizontal  member 
about  D'j  the  vertical  force  at  F  is  600  Ib.  acting  downward. 
Indicate  it  at  once  on  the  sketch;  and  do  likewise  with  each  force, 
as  soon  as  its  magnitude  and  sense  are  found.  The  vertical 
force  at  Df  ,  is  900  Ib.  acting  upward.  Since  the  inclined  mem- 
ber is  pulling  downward  600  Ib.  at  F,  the  horizontal  member 
must  be  pulling  upward  600  Ib.  at  F1.  A  vertical  force  of  600 
Ib.  acts  downward  at  Cr,  and  upward  at  C.  A  vertical  force  of 
900  Ib.  acts  downward  at  D.  Balance  the  moments  of  the 
forces  acting  on  the  inclined  member  about  F'  . 


SEC.  IV       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM 


IOI 


A  horizontal  force  of  450  Ib.  acts  to  the  right  at  C",  and  to  the 
left  at  F';  consequently  a  horizontal  force  of  450  Ib.  acts  to  the 
left  at  C  and  to  the  right  at  F.  Finally  a  horizontal  force  of  450 
Ib.  acts  to  the  left  at  D',  and  to  the  right  at  D. 


*,,,ss,S,sss)$ 
::::;5^:::: 


FIG.  72. 


FIG.  73.[ 


FIG.  75- 


FIG.  76.  FIG.  75. 

Inspect  each  member  and  see  if  the  forces  acting  on  each  are 
in  equilibrium.  Draw  shear  and  bending  moment  diagrams  for 
the  vertical  and  horizontal  members. 


102 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  II 


The  resultant  of  the  two  forces  acting  at  F'  must  have  a 
direction  through  C',  otherwise  the  inclined  member  would 
rotate  and  not  be  in  equilibrium.  Likewise  the  resultant  at  C' 
passes  through  F' ',  and  the  member  resists  a  longitudinal 
stress.  Whenever  forces  act  at  only  two  points  on  a  member, 
the  member  reacts  as  a  tie  or  strut  and  not  as  a  beam. 

64.  Problems. 

1-6.  The  structures  shown  in  Figs.  72  to  77  inclusive  are  to  be  treated  as 
follows:  Determine  the  horizontal  and  vertical  components  of  the  reactions, 
sketch  each  member  separately  and  find  the  horizontal  and  vertical  components 
of  the  forces  acting  thereon. 


/4f ->|<-  —IO'---- 


FIG.  77. 

65.  The  portal  frame  (Fig.  780),  resisting  a  horizontal  force 
of  600  Ib.  at  D  has  four  unknown  magnitudes  represented  in  the 
reactions — i.e.  a  vertical  and  a  horizontal  force  at  each  support; 
but  the  frame,  being  composed  of  two  rigid  bodies  hinged  at  C, 
is  statically  determinate.  Each  vertical  member  or  column  is 
continuous.  The  two  horizontal  members  and  the  inclined 
members  are  hinged  at  C.  Balance  the  moments  of  all  the 
forces  about  A . 

600  X  24 


16 


=  900 


A  vertical  force  of  900  Ib.  acts  upward  at  J3,  consequently  a 
vertical  force  of  900  Ib.  acts  downward  at  A.  Balance  the 
moments  of  the  forces  acting  on  the  right  portion  of  the  struc- 
ture about  C. 

900  X  8 

-  =  240 
30 


SEC.  IV       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM 


103 


A  horizontal  force  of  240  Ib.  acts  to  the  left  at  B.  Balance  the 
moments  of  the  forces  acting  on  the  left  portion  of  the  structure 
about  C. 

c 


600 ' 


FIG.  78 a 


360* 


1200 


960* 


1200*    1200* 


|H 

900* 


t. 

900  * 


FlG.78-b 


240* 


900* 


900  X  8  =  7,200 
600  X  6  =  3,600 


30)10,800 
360 

A  horizontal  force  of  360  Ib.  acts  to  the  left  at  A. 

Each  member  in  the  structure  might  now  be  sketched  sepa- 
rately, and  the  stresses  analyzed  by  the  algebraic  method,  as  in 


104  THEORY    OF    FRAMED    STRUCTURES  CHAP.  II 

Article  63.  In  many  cases,  however,  and  especially  if  the  frame 
is  more  complex,  a  graphic  solution  by  means  of  the  stress 
diagram  is  desirable;  but  before  a  stress  diagram  can  be  drawn, 
the  bending  effect  of  the  columns  upon  the  frame,  caused  by  the 
horizontal  forces  at  A  and  B,  must  be  considered.  Each 
column  is  sketched  separately  as  before  (Fig.  786.)  The  known 
forces  acting  on  the  columns  are  the  reactions  at  A  and  B  and 
the  load  of  600  Ib.  at  D.  Other  forces,  representing  the  action 
of  the  frame  HIJK  on  the  columns,  must  be  supplied  at  E,  D, 
F  and  G.  It  is  necessary  to  consider  only  the  horizontal  forces 
required  for  the  equilibrium  of  the  columns  at  these  points, 
since  the  vertical  forces  in  the  columns  do  not  cause  bending  in 
the  columns.  The  horizontal  forces  necessary  for  the  equilib- 
rium of  the  columns  are  as  follows: 

At  G  24°  *  24  =  960  acting  to  the  left. 
6 

At  F  240  +  960  =  1,200  acting  to  the  right. 

At  E  3  °     —  =  1,440  acting  to  the  left. 

At  D  1,440  +  360  —  600  =  1,200  acting  to  the  right. 

These  four  horizontal  forces  represent  the  bending  action  of 
the  frame  on  the  columns.  The  reaction  of  the  columns  on  the 
frame  are  equal  and  opposite  as  indicated  on  the  frame.  The 
vertical  forces  acting  on  the  columns  at  A  and  B  are  transferred 
to  H  and  K.  The  frame  is  in  equilibrium,  and  a  stress  diagram 
may  be  drawn. 

66.  The  portal  frame  (Fig.  790)  is  statically  indeterminate. 
The  vertical  reaction  at  either  support  may  be  determined  by 
balancing  the  moments  of  all  the  forces  about  the  other  support, 
thereby  eliminating  the  two  horizontal  reactions.  The  hori- 
zontal reactions  cannot  be  found  by  the  principles  of  statics. 
In  order  that  an  approximate  estimate  of  the  stresses  may  be 
made,  current  engineering  practice  assumes  that  the  horizontal 
load  is  supported  equally  by  the  two  columns.  The  two  columns 
are  sketched  separately,  as  before.  The  horizontal  loads  and 
reactions  are  shown  on  the  outside  of  the  columns  and  the  forces, 
interacting  between  columns  and  frame  are  shown  on  the  inside 


SEC.  IV       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM 


105 


of  the  columns.  The  latter  forces  have  been  transferred  to  the 
frame  in  Fig.  79^,  and  the  vertical  reactions  indicated.  A  stress 
diagram  may  now  be  drawn. 


1000 


woo 


1300 


6100 


600* 

opfiTf: 

780C 

^ 

^1000 

/\/\ 

9100 

J,     • 

/i       / 

^o     -* 

7^a<?^                                   7<9(?/ 

$/0<f  /            \/             \  ^/^/ 

Hs 

*—  f 

V—                           /" 

1300 
jrj  — 

3800*     Ft  6.  79  b        3800* 
1300^ 

38 

"T"              "^ 

3^»       r/G.75a          J5 

(^ 

600 


<- 


—4*5*%)'- 


FIG.  8oa. 


1000] 


lOOQl 


PIG.  8o&. 


67.  The  portal  frame  (Fig.  Soa)  is  supported  by  columns 
which  are  fixed  at  their  bases  instead  of  hinged,  and  the 
columns  are  not  free  to  rotate  about  their  points  of  support. 
There  are  three  unknown  quantities  at  each  support — a  hori- 


io6 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  II 


zontal  force,  a  vertical  force  and  a  couple;  for  which  six  equa- 
tions or  statements  are  necessary  for  a  solution.  The  principles 
of  statics  furnish  three  of  these  equations,  which  are  to  be 
augmented  by  three  statements  from  which  the  three  additional 
equations  are  obtained.  The  first  supplementary  equation  is 
deduced  from  the  same  assumption  that  was  made  in  Article  66, 
viz.,  that  each  horizontal  reaction  is  %  X  2,600  =  1,300  Ib. 
An  exaggerated  idea  of  the  distortion  which  the  structure 
undergoes  when  loaded,  is  illustrated  in  Fig.  Sob.  If  the  columns 
are  rigidly  fixed  at  their  bases,  a  tangent  to  the  axis  of  each 
column  remains  vertical  after  the  column  is  bent.  The  dif- 

600 


woo 


PIG.  8oc. 


FIG.  Sod. 


ference  in  the  displacements  d\  and  d%  is  assumed  negligible,  when 
compared  with  either  d\  or  d2.  A  similar  assumption  applies  to 
the  points  E  and  F,  and  the  frame  CDEF,  although  displaced 
from  its  original  position,  is  assumed  to  remain  a  rectangle. 
On  the  basis  of  this  assumption  the  point  of  contraflexure, 
or  zero  bending  moment,  in  each  column  is  midway  between  the 
base  of  the  column  and  the  bottom  chord  of  the  frame.  This  is 
another  way  of  saying  that  the  columns  may  be  considered 
hinged  at  G  and  /. 

The  lower  portions  of  the  columns  are  shown  in  Fig.  Soc. 
The  horizontal  displacements  of  G'  and  /'are  grossly  exaggerated 
in  the  figure.  The  couple  formed  by  the  two  vertical  forces  is 
so  very  small  when  compared  to  the  horizontal  couple,  that  it 
may  be  neglected  without  appreciable  error.  Balance  the 
moments  of  all  the  forces  about  G' . 

MI  =  15  X  1,300  =  19,500  ft.-lb.  counter-clockwise. 


SEC.  IV       APPLICATION    OF   PRINCIPLES    OF   EQUILIBRIUM 


107 


A  horizontal  force  of  1,300  Ib.  acting  to  the  right,  and  a  vertical 
force  Vi  acting  upward  must  be  supplied  at  G'  by  the  upper  part 
of  the  column,  in  order  that  the  lower  part  may  be  in  equilib- 
rium. The  forces  required  at  /'  (Fig.  8oc)  are  likewise 
indicated. 

M 2  =  15  X  1,300  =  19,500  ft.-lb.  counter-clockwise. 
The  forces  at  G'  and  /'  (Fig.  8oc)  are  shown  reversed  at  G  and  J 
(Fig.  Sod)  and  the  solution  may  proceed  as  in  Article  66. 

Structures  of  this  character  are  seldom,  if  ever,  designed  with 
pinned  connections  at  the  column  bases,  as  illustrated  in  Fig. 
790.  It  is  equally  true  that  the  column  connections  to  the 
foundation  are  seldom,  if  ever,  made  with  sufficient  rigidity  to 
justify  the  assumption  of  zero  bending  moment  at  one-half  the 
distance  from  the  foot  of  the  column  to  the  frame,  as  shown  in 
Fig.  806.  The  point  of  zero  bending  moment  in  nearly  all  cases 
lies  somewhere  between  these  two  extremes,  and  is  generally 
assumed  to  be  one-third  the  distance  from  the  foot  of  the 
column  to  the  first  connection  with  the  frame.  This  assump- 
tion makes  h  =  20  ft.  in  Fig.  Sod. 


<- 8®  6=48  -- 


FIG.  81. 

68.  Problems. 

1.  Complete  the  solution  for  Fig.  796  and  the  two  solutions  in  Fig.  8od,  where 
h  =  15  ft.  in  one  instance  and  h  =  20  ft.  in  the  other.     Compare  the  results  of 
the  three  solutions. 

2.  Draw  a  stress  diagram  for  the  Fink  truss  in  Fig.  81,  assuming  that  the 
point  of  contraflexure  in  each  column  is  one-third  the  distance  from  the  base 
of  the  column  to  the  foot  of  the  knee  brace. 


CHAPTER  III 
ROOF  TRUSSES 

SEC.  I.    TYPES  AND  LOADS 

69.  Standard  Types. — Roof  trusses  have  been  built  in  a 
great  variety  of  forms  in  conformance  with  architectural 
requirements,  but  under  ordinary  conditions  the  form  of  the 
truss  is  determined  by  the  length  of  span,  slope  of  roof  and 
the  clear  head  room  underneath.  The  most  common  types 
used  in  current  practice  are  shown  in  Fig.  82. 

The  Howe  truss  is  especially  adapted  to  wood  construction. 
The  top  chord  and  diagonals  are  wood,  the  vertical  tension 
members  are  steel  rods  and  the  bottom  chord  is  either  wood 
or  steel.  This  type  is  also  used  when  the  truss  is  made  entirely 
of  steel. 

The  Fink  truss  is  the  most  common  form  in  use  for  support- 
ing roofs  on  which  the  character  of  the  covering  makes  a  steep 
slope  desirable,  as  in  the  following  cases:  "Bonanza"  tile, 
5  in.  per  foot;  corrugated  sheet  steel,  6  in.  per  foot;  slate, 
7  in.  per  foot.  The  advantages  of  this  type  are  the  short 
compression  members,  which  make  for  economy;  and  the 
variations  in  number  and  length  of  panels  in  the  top  chord 
for  any  given  length.  When  the  short  member  in  a  Fink 
truss  is  replaced  by  two  members,  the  structure  is  sometimes 
called  a  fan  truss. 

The  Warren  truss  is  used  extensively  for  tar  and  gravel  roofs 
with  the  following  slopes:  %  in.  per  foot  for  spans  below  40 
ft.  in  length,  i)4  in.  per  foot  for  spans  60  ft.  and  over,  and  i 
in.  per  foot  for  intermediate  spans. 

Saw-tooth  trusses  have  top  chords  of  unequal  length.  Win- 
dows are  placed  in  the  plane  of  the  shorter  legs,  which  face 
north  in  the  northern  hemisphere  to  avoid  the  direct  rays  of 
the  sun. 

108 


SEC.  I 


ROOF    TRUSSES 


109 


Three-hinged  arches  are  more  economical  than  Fink  or  Warren 
trusses  for  spans  above  about  125  ft.  in  length. 

Monitor  trusses  are  auxiliary  frames  which  may  be  added  to 
the  main  trusses  to  provide  light  or  ventilation. 

A  bent  consists  of  a  roof  truss  and  the  columns  which  support  it. 


Howe 


Fink 


Fan 


Warren 
v/ifh  moni+or 


Saw  Tooth 


FIG.  82. 

A  bay  is  the  space  between  adjacent  trusses,  and  the  distance 
center-to-center  of  trusses  is  called  the  bay  length. 

70.  Purlins. — The  roof  covering  is  supported  by  purlins 
which  are  carried  by  the  trusses.  The  purlins  are  usually 
channels,  although  I-beams  are  sometimes  used.  It  was 
formerly  the  custom  to  lay  out  the  design  so  that  the  purlins 
would  rest  on  the  top  chord  of  the  truss  at  the  panel  points; 
but  current  practice  seems  to  disregard  this  feature  by  fre- 


110  THEORY    OF   FRAMED    STRUCTURES  CHAP.  Ill 

quently  making  the  panel  lengths  longer  than  the  purlin 
spacing.  This  arrangement  reduces  the  number  of  web 
members  otherwise  required;  but  increases  the  size  of  the  top 
chord,  which  is  longer  between  panel  points  and  must  also 
perform  the  additional  duty  of  a  beam  in  supporting  the 
purlins  between  panel  points.  The  extra  weight  in  the  top 
chord  more  than  offsets  the  weight  saved  in  the  web  members, 
and  thus  adds  weight  to  the  structure;  but  the  cost  is  less  on 
account  of  time  saved  in  fabrication. 

71.  The  bracing  is  an  important  part  of  any  structure  and 
is  too  frequently  given  little  attention.  It  plays  a  leading 
role  during  erection,  especially  if  the  trusses  are  long  or  the 
building  is  high;  but  the  chief  purpose  of  bracing  is  to  carry 
the  wind  loads  and  vibratory  effects  of  machinery  and  the 
like  to  the  foundations.  To  provide  for  wind  pressure  on  the 
ends  of  a  building,  bracing  is  placed  in  the  planes  of  the  top 
and  bottom  chords  of  the  trusses,  which  transfers  the  wind 
load  to  the  eaves.  If  the  trusses  are  supported  on  columns, 
the  load  is  transferred  from  the  eaves  to  the  ground  by  bracing 
between  the  side  columns.  This  bracing  is  placed  in  the  end 
bays  and  in  approximately  every  fourth  bay. 

The  wind  load  on  the  sides  of  a  building  supported  on  columns 
may  be  taken  to  the  foundation  by  two  different  paths.  Each 
truss  and  the  columns  supporting  it  may  act  as  a  unit,  in  which 
case  the  wind  load  on  each  adjacent  half  bay  is  carried  to  the 
ground  through  the  columns  acting  also  as  a  beam.  This 
device  is  used  whenever  possible,  for  it  is  a  sound  economic 
principle  to  transfer  a  load  to  the  ground  over  the  shortest 
path  possible.  Whenever  the  building  contains  a  crane  run- 
way, the  use  of  knee  braces  from  the  bottom  chord  to  columns 
is  often  objectionable  or  impossible.  If  sufficiently  rigid 
connections  between  the  trusses  and  columns  cannot  be  made, 
the  wind  and  transverse  thrust  from  the  crane  may  be  taken 
to  the  ends  of  the  building  by  diagonal  bracing  in  every  bay 
in  the  plane  of  the  bottom  chords  of  the  trusses.  At  the  ends 
of  the  building  the  load  is  transferred  to  the  ground  by  bracing 
between  the  end  columns.  This  method  of  bracing  is  not  effec- 
tive in  a  building  so  long  that  expansion  joints  are  necessary. 


SEC.  I  ROOF    TRUSSES  III 

The  uncertainty  concerning  the  nature  and  amount  of  wind 
load  (as  will  be  shown  later)  and  the  vibratory  action  of 
machinery;  and  the  fact  that  these  loads  may  often  be  carried 
to  the  ground  over  more  than  one  path,  makes  the  problem  of 
estimating  the  probable  resulting  stresses  in  the  bracing  a 
very  difficult  one.  As  a  matter  of  fact  it  is  seldom  attempted 
except  in  a  structure  of  extraordinary  capacity.  The  design 
of  bracing  is  governed  more  often  by  the  lengths  of  its  members 
than  by  the  stresses  they  are  supposed  to  carry. 

72.  Spacing  of  Trusses. — The  economic  bay  length  is  a  func- 
tion of  several  variables — the  cost  of  the  purlins,  trusses,  col- 
umns and  foundations,  all  enter  into  the  problem.     If  the  bay 
length  is  from  8  to  12  ft.,  2  or  3  in.  tongue  and  grooved  planking, 
resting  directly  on  the  trusses  may  be  used,  in  which  case  pur- 
lins are  not  required.     When  purlins  are  used,  the  bay  length 
is  usually  about  16  ft.  for  spans  up  to  65  ft.     For  longer  spans 
the  bay  length  is  about  one-quarter  of  the  span. 

73.  The  dead  loads  supported  by  a  truss  include  the  roof 
covering,   the  purlins,   the  bracing,   the  weight  of  the  truss 
itself  and  any  stationary  loads,  such  as  a  machinery  floor  or 
ceiling  which  may  be  suspended  from  the  truss. 

Roof  Covering. — The  character  and  weight  of  the  roof  cover- 
ing will  influence  the  size  and  spacing  of  the  purlins. 

The  weights  per  square  foot  of  roof  surface  for  the  more  com- 
mon materials  are  as  follows:  shingles,  2  to  3  lb.;  slate,  8  to 
10  lb. ;  corrugated  steel,  i  to  3  lb. ;  tile,  8  to  25  lb. ;  felt  and  gravel, 
8  to  10  lb.;  boards  i  in.  thick,  3  to  4  lb. 

The  purlins  support  the  roof  covering,  the  live  load  (wind  or 
snow  or  both),  and  their  own  weight.  Purlins  are  assumed  to 
weigh  about  3  lb.  per  square  foot  of  the  roof  surface.  This 
assumption  should  be  compared  with  the  actual  weight  after 
the  purlins  have  been  designed.  If  the  assumed  and  actual 
weights  differ  materially,  a  correction  should  be  made,  and  the 
design  modified  if  necessary. 

Bracing. — The  weight  of  the  bracing  is  a  relatively  small  item 
and  is  generally  not  considered,  in  estimating  the  total  dead 
load.  Whenever  considered,  it  is  usually  estimated  at  about 
i  lb.  per  square  foot  of  horizontal  surface. 


112 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  Ill 


Trusses. — The  actual  weight  of  a  truss  cannot  be  determined 
until  the  design  and  shop  drawings  have  been  made.  It  is 
necessary,  however,  to  have  an  approximate  weight  of  the 
truss  when  estimating  the  total  dead  load.  The  weight  of 
a  steel  truss  is  a  function  of  several  variable  quantities,  i.e.  the 
length  of  span,  the  length  of  bay,  the  load  to  be  supported,  the 
specifications  used  in  the  design,  the  type  of  truss,  the  slope  of 
the  roof,  the  position  of  the  purlins  with  reference  to  the  panel 
points,  and  the  amount  of  unavoidable  excess  metal  necessary 
in  order  that  the  design  may  be  in  accordance  with  the  specifica- 
tions. Empirical  formulas,  frequently  used  for  finding  the 
weights  of  trusses,  are  as  a  rule  not  trustworthy;  for  they  include 
only  a  few  of  the  determining  factors,  and  the  limits  within 
which  these  factors  are  applicable  are  not  specified.  The 
approximate  weights  given  in  the  following  table  may  be  used 
for  the  preliminary  design.  After  the  stresses  have  been  found 
and  the  design  made,  a  closer  approximation  to  the  actual 
weight  may  be  made  in  the  following  manner.  Find  the  weight 
of  the  members  in  the  truss,  taking  the  distance  center  to  center 
of  panel  points  for  the  length  of  each  member;  and  add  25  per 
cent  for  details.  If  the  approximate  weight  of  the  truss  thus 
found  differs  from  the  assumed  weight  by  an  amount  sufficient 
to  change  the  total  load  (dead  and  live)  5  per  cent,  the  designer 
should  consider  the  desirability  of  revising  the  design. 

TABLE  I. — WEIGHTS  OF  ROOF  TRUSSES  IN  POUNDS 


Total  load  in  pounds  per  linear  foot  of  span 

Span, 

feet 

500 

600 

700 

800 

900 

1,000 

30 

850 

950 

I  ,OOO 

1,100 

1,150 

1,350 

40 

1,250 

1,500 

1,650 

1,  800 

1,950 

2,IOO 

50 

2,000 

2,2OO 

2,350 

2,650 

3,000 

3,300 

60 

2,350 

2,800 

3,200 

3,500 

3,800 

3,900 

70 

3,300 

3,850 

4,100 

4,600 

4,950 

5,350 

80 

4,2OO 

4,850 

5,350 

6,000 

6,250 

6,900 

74.  The  snow  load  varies  with  the  latitude,  humidity  and 
slope  of  roof.     The  weight  of  freshly  fallen  snow  varies  from  5 


SEC.  I  ROOF    TRUSSES  113 

to  12  lb.  per  cubic  foot,  according  to  its  dampness.  The  weight 
when  packed  may  vary  from  15  to  40  lb.  per  cubic  foot.  Snow 
is  seldom  considered,  if  the  roof  has  an  inclination  of  45°  or 
more  to  the  horizontal.  Specifications  usually  allow  the  follow- 
ing weights  per  horizontal  square  foot  for  snow:  New  England, 
30  lb.;  New  York  and  Chicago,  20  lb.;  Baltimore,  Cincinnati 
and  St.  Louis,  10  lb. 

75.  The  Wind  Load.1 — The  effect  of  wind  upon  structures  is 
at  present  an  unsolved  problem.  Nearly  all  scientists  since 
Newton's  time  agree  that  wind  pressures  of  equal  densities  vary 
as  the  square  of  the  velocity.  This  law  may  be  expressed  by 
the  empirical  equation 

P  =  kV\ 

where  P  represents  the  pressure  in  pounds  per  square  foot  on  a 
surface  normal  to  the  wind,  V  is  the  wind  velocity  in  miles  per 
hour  and  k  is  a  constant  factor  to  be  determined  by  experiment. 
Scientists  do  not  agree  on  the  value  of  k,  but  current  practice 
seems  to  consider  0.004  as  a  proper  value;  whence 

P  =  o.oo4F2. 

If  this  formula  is  accepted,  the  following  wind  pressures  on 
a  normal  surface  will  result: 

WIND  VELOCITIES  PRESSURE,  NORMAL  SURFACE, 

MILES  PER  POUNDS  PER  SQUARE 

HOUR  FOOT 

60  14.4 

70  19.6 

80  25.6 

90  32.4 

100  40 .  o 

Again  authorities  differ  as  to  the  maximum  velocity  for  which 
provision  should  be  made.  It  is  considered  useless  to  design  a 
structure  sufficiently  strong  to  resist  a  tornado  which  may  come 
once  in  a  lifetime  and  be  of  short  duration.  High  velocities 
come  in  gusts  and  act  over  relatively  small  areas.  In  the  light 
of  our  present  knowledge,  the  provision  for  a  wind  pressure  of  20 
lb.  per  square  foot  on  vertical  surfaces  seems  ample. 

The  action  of  wind  upon  an  inclined  surface  is  also  a  per- 

lThe  question  of  wind  pressure  on  buildings  has  been  presented  in  an 
admirable  manner  by  R.  FLEMING,  in  the  Engineering  News,  Jan.  28  and 
Feb.  4,  1915. 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  Ill 


plexing  problem  to  which  scientists  have  given  much  attention. 
It  is  not  correct  to  consider  the  effect  of  the  wind  on  an  inclined 
roof  surface  to  be  the  same  as  upon  a  vertical  surface;  nor  is  it 
correct  to  resolve  the  horizontal  force  of  the  wind  into  two 
components,  the  one  normal  to  the  roof,  and  the  other  along 
the  surface  of  the  roof;  for  the  effect  of  the  normal  component 
only,  would  be  resisted  by  the  roof  covering. 

In  1829  Duchemin,  a  French  army  officer,  investigated  the 
wind  pressure  on  inclined  surfaces;  and  his  conclusions,  approxi- 
mately  verified    by   Langley  in    1888,   have  been  generally 
accepted.     Duchemin's  empirical  formula  is: 
2  sin  A 

•L  n   —    *  .         .     9     A 

i  +  sin2  A 

in  which  Pn  =  pressure  normal  to  the  roof, 
P  =  pressure  on  a  vertical  surface, 
A  =  angle  of  inclination  to  the  horizontal. 

When  A  =  45°  or  more,  Pn  should  be  taken  equal  to  P.  The 
normal  pressure  for  various  slopes  and  angles  of  inclination 
are  as  follows: 

TABLE  II.— NORMAL  WIND  PRESSURES  ON  ROOF  SURFACES 


Angle  of  inclination  or  slope 
A 

Slope 

Normal   pressures   in   pounds 
square    foot     when     pressure 
vertical  surface  is  P  —  20  Ib. 
square  foot 

per 
on 
per 

4°  46' 

i  in.  to  i  ft. 

3-3 

5°    o' 

3-5 

9°  28' 

2  in.  to  i  ft. 

6-4 

10°      0' 

6.8 

14°      2' 

3  in.  to  i  ft. 

9-2 

iS°    o' 

9.6 

18°  26' 

4  in.  to  i  ft. 

ii-5 

20°     0 

12.  2 

22°  37' 

5  in.  to  i  ft. 

13-4 

25°    o' 

14.4 

26°  34' 

6  in.  to  i  ft. 

14.9 

30°    o' 

16.0 

30°  15' 

7  in.  to  i  ft. 

16.1 

33°  4i' 

8  in.  to  i  ft. 

17.0 

35°    o' 

17-3 

40°    o' 

18.2 

45°  90' 

20.0 

SEC.  II  ROOF   TRUSSES 


SEC.  II.     STRESS  ANALYSIS  FOR  WIND 

76.  Wall  Bearing  Trusses. — The  stresses  in  a  roof  truss  are 
usually  determined  by  graphic  methods.  Theoretically,  two 
stress  diagrams  are  required — one  for  vertical  loads,  and  one 
for  the  wind  loads  acting  normal  to  the  roof.  In  practice, 
however,  a  separate  stress  diagram  for  wind  is  seldom  drawn, 
except  for  trusses  supported  on  columns;  and  then  only  for 
localities  having  high  winds  and  no  snow.  The  reason  for  this 


(b) 


FIG.  83. 

will  now  be  shown  in  connection  with  the  Fink  truss  in  Fig. 
830.  The  following  data  will  be  used:  The  truss  is  supported 
on  walls;  span,  40  ft.;  rise,  10  ft.;  length  of  bay,  15  ft.;  wind 
pressure,  20  Ib.  per  square  foot  on  a  vertical  surface.  The 
normal  wind  pressure,  taken  from  the  table  in  Article  75,  is 
14.9  Ib.  per  square  foot  of  roof  surface.  The  length  of  the  top 
chord  is  22.36  ft.;  hence,  the  area  subject  to  wind  pressure 
which  each  truss  supports  is  22.36  X  15  =  334-8  sq.  ft.;  and  the 
total  wind  load  is  334.8  X  14.9  =  5,000  Ib.;  or  1,250  Ib.  per 
panel.  A  half-panel  load  is  supported  at  the  eaves  and  peak, 
and  a  full  panel  load  at  each  of  the  three  other  panel  points. 


Il6  THEORY    OF   FRAMED    STRUCTURES  CHAP.  Ill 

The   horizontal  and  vertical  components  of  the  total  load 
are  2,236  and  4,472  Ib.  respectively.    Let  V\,  HI,  V2  and  H2 
represent,  respectively,  the  vertical  and  horizontal  components 
of  the  left  and  right  reaction  (see  Article  52)  . 
F2=  5,000  X  11.18  = 

Vi  =  4,472  "  i,398  =  3^74 

Since  the  horizontal  reactions  are  statically  indeterminate, 
four  assumptions  as  to  the  values  of  HI  and  Hz  will  be  con- 
sidered. 

Case  I.  —  The  total  horizontal  reaction  is  carried  by  the  left 
support. 

Hi  =  2,236 
#2  =  0 

Case  II.  —  The  total  horizontal  reaction  is  carried  by  the 
right  support 

#1  =  0 
Hz  =  2,236 
Case  III.  —  The  horizontal  reactions  are  equal 

H!  =  H2  =  1,118 

Case  IV.  —  The  horizontal  reaction  at  each  support  is  pro- 
portional to  the  vertical  reaction 

x  2'236  =  I>537 


The  stresses  for  all  four  cases,  determined  by  the  stress 
diagram  in  Fig.  83  b,  are  recorded  in  Table  IV;  where  each 
member  is  numbered  to  correspond  with  Fig.  87.  The  bottom 
chord  is  straight,  and  in  line  with  the  two  points  of  support;  and 
because  of  this,  the  stress  in  any  member,  except  those  of  the 
bottom  chord,  is  the  same  for  the  four  different  assumptions 
regarding  the  horizontal  reactions. 

77.  Trusses  Supported  on  Columns.  —  The  Mill  Building 
Bent.  —  When  the  truss  of  Fig.  83  a  is  supported  by  columns,  as 
shown  in  Fig.  84;  six  unknown  quantities  are  involved  in  the 
reactions  —  a  vertical  force,  a  horizontal  force  and  a  resisting 


SEC.  II 


ROOF    TRUSSES 


moment  at  each  support.  The  problem  is  therefore  stati- 
cally indeterminate  in  the  third  degree,  and  the  three  static 
equations  of  equilibrium  must  be  augmented  by  three  elastic 
equations,  in  making  an  exact  analysis  of  the  reactions.  The 
derivation  of  these  elastic  equations  is  a  long  and  involved 
process,  and  will  not  be  considered  here.  In  place  of  the 
elastic  equations,  three  assumptions  are  made.  Hitherto,  it 
has  been  the  practice  to  treat  the  mill  building  bent  as  a  portal 
frame,  the  same  assumptions  being  made  as  outlined  at  the 
end  of  Article  67;  viz.,  that  the  point  of  contraflexure,  or  zero 
bending  moment,  in  each  column  is  one-third  of  the  distance 
from  the  base  of  the  column  to  the  foot  of  the  knee-brace;  and 
the  total  horizontal  shear  is  equally  resisted  by  each  column. 


T 


^  \A      V                     V      A/   \ 

Y 

7A        A\ 

A 

-i 

r 
f. 

; 

~r  f 

i 

X 

y 

V 

v  > 

! 

FIG.  84. 

A  recent  investigation,1  made  by  Mr.  S.  R.  Oflutt  at  the 
University  of  Illinois,  under  the  direction  of  the  author  has 
proved  that  the  assumption  of  equal  horizontal  reaction  is  no 
longer  justified. 

Twenty  structures  having  the  general  outline,  as  shown  in 
Fig.  84,  and  dimensions  given  in  Table  III  were  designed  and 
investigated.  Fink  trusses,  having  a  slope  of  6  in.  to  i  ft., 
were  used  in  all  cases.  Each  truss  had  eight  panels,  except 
the  2o-ft.  trusses,  which  had  four  panels.  The  length  of  the 
bay  was  15  ft.  The  truss  for  each  span  length  /  was  designed 
for  a  uniform  vertical  load  of  about  40  Ib.  per  square  foot  of 
horizontal  surface.  This  load  was  varied  somewhat  in  several 

1This  investigation  is  to  be  published  as  a  Bulletin  of  the  University  of 
Illinois  Engineering  Experiment  Station. 


n8 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  Ill 


instances  to  note  the  effect  upon  the  final  results.  The  mem- 
bers A  were  made  about  twice  as  large  as  was  required  for  the 
vertical  load.  The  knee-braces  were  designed  for  a  wind 
pressure  of  20  Ib.  per  square  foot  on  the  vertical  height  h,  and 
14.9  Ib.  per  square  foot  of  roof  surface  normal  to  the  roof.  The 
columns  in  general  were  designed  to  support  their  vertical  load 
and  the  additional  effect  of  bending;  by  assuming  that  the  point 
of  contraflexure  was  one-third  the  distance  from  the  base  to  the 
knee-brace;  and  that  the  horizontal  reactions  were  equal. 


TABLE  III. — RATIOS  FOR  HORIZONTAL  REACTIONS  AND  POINTS  OF  CONTRA- 
FLEXURE IN  MILL  BUILDING  COLUMNS 


Span 
length 

Column 
height 

Ratio 

Pin-con- 
nected 

Rigidly  fixed 

h 

I 

HR 

a* 

X 

y 

n 

h 

H 

H 

d 

d 

i 

2 

3 

4 

5 

6 

7 

20 

10 

0.500 

0.399 

0.385 

0.518 

0.654 

20 

12 

0.600 

0-379 

o-354 

0.486 

0.620 

20 

16 

0.800 

0-358 

0.323 

0-435 

0.585 

2O 

i9 

0.950 

0.356 

0.309 

0-439 

0.569 

30 

16 

0-533 

0.380 

0-353 

0.481 

0-593 

30 

21 

0.700 

0.364 

0.324 

0-435 

0.570 

30 

26 

0.867 

0-354 

0.309 

0.432 

0-547 

30 

31 

1-033 

0-340 

o.  290 

0-439 

0.562 

40 

16 

0.400 

0.410 

0.401 

0.621 

0.612 

40 

21 

0-525 

0.385 

0.356 

0.488 

0.581 

40 

26 

0.650 

0-373 

0-335 

0.471 

0.560 

40 

31 

0-775 

0.358 

0.314 

0-457 

0.568 

50 

16 

0.320 

0.425 

0.429 

0-634 

0.601 

50 

21 

0.420 

0-397 

0-375 

0.498 

o.57i 

50 

26 

0.520 

0.382 

0-349 

0-465 

0-555 

50 

31 

0.620 

0.369 

0.332 

0.452 

o.558 

60 

16 

o.  267 

0-439 

0-453 

o  .  643 

0-593 

60 

21 

0.350 

0.407 

0.388 

0-494 

0-574 

60 

26 

0-433 

0.390 

0.361 

0-454 

0-558 

60 

31 

0.517 

0-379 

0-345 

0.429 

0-554 

SEC.  II  ROOF   TRUSSES  119 

This  column  design  also  was  varied  in  several  instances  to  note 
the  general  effect.  Each  bent  thus  designed  was  then  analyzed, 
strictly  in  accordance  with  the  elastic  theory  of  structures 
by  the  method  of  deflections.  This  theory  is  too  complicated 
to  be  explained  here.  One  assumption,  not  strictly  in  accord 
with  actual  conditions  was  made;  viz. — that  the  truss  and  knee- 
brace  members  were  pin-connected..  The  horizontal  wind 
loads  were  concentrated  at  points  about  5  ft.  apart,  along  the 
column.  The  wind  loads  normal  to  the  roof  were  concen- 
trated at  the  panel  points  of  the  truss.  The  intensity  of 
the  roof  loads  was  taken  in  accordance  with  Duchemin's 
formula.  The  reader  is  cautioned  against  drawing  too  general 
conclusions  from  the  data  given  in  Table  III ;  for  only  a  limited 
number  of  structures  was  analyzed,  and  all  trusses  were  of  the 
same  type  and  slope.  The  analysis  was  made  for  columns 
pin-connected  at  their  bases,  and  rigidly  fixed  at  their  bases. 
The  results  given  in  Table  III  are  ratios,  and  are  thus  explained. 

H  —  wind  load  on  windward  or  left  column,  exclusive  of  the 

half-panel  load  at  the  base,  plus  horizontal  component 

of  the  normal  roof  load. 

HR  =  horizontal  reaction  at  base  of  leeward  or  right  column. 
d  =  distance  from  base  of  column  to  foot  of  knee-brace. 
x  =  distance  from  base  of  column  to  point  of  contraflexure 

of  windward  column. 
y  =  distance  from  base  of  column  to  point  of  contraflexure 

of  leeward  column. 

Suppose  that  H  =  10,000  Ib.  for  the  structure,  in  which 
/  =  40  ft.  and  h  =  21  it.  If  the  columns  are  pin-connected,  the 
horizontal  reaction  at  the  base  of  the  leeward  column  is 

HR  =  10,000  X  0.385  =  3,850  Ib. 
If  the  columns  are  fixed  and  d  =  15,  then 

HB  =  10,000  X  0.356  =  3,560  Ib. 
*  =  15  X  0.488  =  7.32  ft. 
y  =  15  X  0.581  =  8.71  ft. 

The  structure  is  statically  determinate  when  one  horizontal 
reaction  and  the  points  of  contraflexure  are  known. 


120 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  Ill 


The  columns  are  seldom,  if  ever,  built  with  pin  connections; 
nor  are  they  sufficiently  anchored  to  the  masonry  to  insure  a 
perfectly  rigid  connection.  The  actual  conditions  existing 


— — 4o'-o- 


2920 


1500 


s*o   D>>f  \74\/             V 

10  V>s^  ^57  ^55 

*      KMC\\/                  A 
b_^  4576  IX 

vir 

\J77,<?7   77,57 

i<    "•              ""•">• 

4576        B  f 

T/f 

/»o                    (c) 

3?50 

+—4S66 

;??(?, 


3252 


FIG.  85. 


in  all  mill-building  bents  are  to  be  found  between  these  two 
limits.  For  example,  in  the  structure  before  mentioned, 
where  /  =  40  ft.  and  h  =  21  ft.,  the  horizontal  reaction  of  the 
leeward  column  will  be  between  0.385  and  0.356  of  the  hori- 


SEC.  II 


ROOF    TRUSSES 


121 


zontal  load.  The  point  of  contraflexure,  or  point  of  zero 
moment,  in  the  leeward  column  will  be  found  between  the  base 
of  the  column  and  the  point  0.488  of  the  distance  to  the  foot  of 
the  knee-brace.  A  comparison  of  the  data  in  columns  4  and  5 
of  Table  III  indicates  that  0.3  7  5#,  instead  of  0.5^,  is  the 
more  reasonable  estimate  of  the  horizontal  reaction  of  the 
leeward  column;  when  the  columns  are  considered  partially 
fixed.  The  data  given  in  columns  6  and  7  indicates  that  the 
point  of  contraflexure  in  each  column  may  be  taken  at  one-third 


0  2000         4000       6000  Lbs, 


FIG.  86. 

the  distance  from  the  base  of  the  column  to  the  foot  of  the 
knee-brace.  This  conclusion  is  justified  only  when  the  masonry 
is  sufficiently  massive  to  resist  the  moment  thereby  alloted  to  it. 
The  wind  concentrations  shown  in  Fig.  850  were  determined 
from  the  data  given  in  Article  76.  The  sum  of  the  horizontal 
loads,  exclusive  of  the  half-panel  load  of  750  lb.,  is 

H  =  7,786 

If  the  horizontal  reaction  of  the  leeward  column  is  taken  as  three- 
eighths  of  the  total  load  H;  then 

HR  =  0.375  X  7,786  =  2,920 
HL  =  7,786  —  2,920  =  4,866 

The  point  of  contraflexure  in  each  column  will  be  taken  one- 
third  the  distance  from  the  column  base  to  the  foot  of  the  knee- 
brace;  and  the  structure  is  thereby  transformed  into  the 


122  THEORY    OF    FRAMED    STRUCTURES  CHAP.  Ill 

statically  determinate  frame  as  shown  in  Fig.  856,  in  which  the 
moments  are  zero  at  A  and  B. 

Moments  about  A  Moments  about  B 

i,5ooX    5  =      7,5oo  4,472X30=                 134,160 

1,650  X  10  =     16,500  2,236  X  21  =  46,956 

900  X  16  =    14,400  900  X  16  =  14,400 

2,236X21=    46,956  1,650X10=16,500 

4,472X10=    44,720  i,5ooX    5=    7>5o°      85,356 

40)130,086  40)48,804 

3,252  =  VR  1,220  =   VL 

The  truss  and  columns  are  shown  separately  in  Fig.  85^;. 
The  known  forces  acting  on  the  columns  are  shown  by  full  lines. 
The  horizontal  forces,  represented  by  dotted  lines  at  #,  bj  c 
and  dj  which  the  truss  must  exert  on  the  columns  in  order  to  hold 
them  in  equilibrium,  are  determined  as  follows: 

Moments  about  c 

force  at  d  =  -    2— — --  =  7,787  acting  to  the  right, 
force  at  c  =  7,787  —  2,920  =  4,867  acting  to  the  left. 

Moments  about  a 
4,866  X  16  =  77,856 

1,650  X    6  =    9,900 
1,500  X  ii  =  16,500 
1,500  X  1 6  =  24,000     50,400 
6)27,456 

4,576  =  force  at  b  acting  to  the  right. 

force  at  a  =  900  +  1,650  +  4,576  +  1,500  +  1,500  —  4,866  = 

5,260  acting  to  the  left. 


PIG.  87. 


SEC.  II  ROOF   TRUSSES  123 

TABLE  IV. — COMPARISON  OF  WIND  STRESSES — I,OOO-LB.  UNITS 


Mem- 
ber 

Wall  bearing  truss 

Mill      Wind  load 
build-        acting 
ing        vertically, 
truss     both  sides 
Case  V     Case  VI 

Case  I 

Hi  =  2,236 
Hz  =          o 

Case  II 
Hi  =         o 
H2  =  2,236 

Case  III 

//i  =  1,118 
H2  =  1,118 

Case  IV 
Hi  =  i,537 
H2  =      699 

i 

-5-6 

-5-6 

-5-6 

-5-6 

-H-3 
+  9-3 

-9.8 

2 

-5-6 

~5-6 

-5-6 

-5-6 

-U-3 
+  9-3 

-9-3 

3 

-5-6 

-5-6 

-5-6 

-5-6 

-  6.4 
+   LI 

-8.7 

4 

-5-6 

-5-6 

-5-6 

-5-6 

-  6.4 
+   LI 

-8.2 

5 

+  7-0 

+  2.8 

+4-8 

+  5-9 

+6.3 

+  4-6 
-  3-5 

+8.8 

6 

+5-6 

+  2.8 

+3-4 

+4-5 

+4-9 

+  4-5 

-   5-7 

+  7-6 

7 

+  2.8 

+0.6 

+  1-7 

+  2.1 

—   i.i 

+  5-0 
+  1-3 

8 

+  1.4 

+  1.4 

+  1.4 

+  1-4 

+  6.9 
-  9-3 

9 

-2.5 

-2.5 

-2.5 

-2.5 

-   5-0 
+  4-2 

-2.3 

10 

+  2.8 

+  2.8 

+  2.8 

+  2.8 

+   5-5 
-   4-7 

+  2.5 

ii 

+4-2 

+4-2 

+4-2 

+4-2 

+  6.9 

-   4-7 

+3-8 

12 

0 

o 

o 

o 

+  6.3 
—  10.8 

o 

13 

+1.4 

+  1.4 

+  1.4 

(+i-4 

+   1-4 

+  1-3 

14 

—  1-3 

-i-3 

-1-3 

-i-3 

-   i-3 

—  i.i 

124  THEORY    OF   FRAMED    STRUCTURES  CHAP.  Ill 

Since  the  forces  which  the  truss  exerts  on  the  columns  must  be 
respectively  equal  in  magnitude,  but  opposite  in  sense,  to  the 
forces  which  the  columns  exert  on  the  truss;  the  forces  at  a,  b,  c 
and  d  are  therefore  shown  reversed  on  the  truss.  These 
forces  virtually  represent  the  bending  effect  imposed  by  the 
columns  on  the  truss.  The  vertical  reactions,  or  the  thrust  of 
the  columns  on  the  truss,  are  also  transferred  to  the  truss. 
The  truss  is  now  in  equilibrium  under  the  action  of  the  forces 
shown  thereon,  and  a  stress  diagram  may  be  drawn  as  shown  in 
Fig.  86.  The  stresses  are  recorded  as  Case  V  in  Table  IV. 
The  members  are  numbered  to  correspond  with  Fig.  87. 

78.  Stress  Diagram  for  Wind  Unnecessary. — The  stresses 
for  Cases  I  to  V  inclusive  (as  recorded  in  Table  IV)  were 
determined  for  a  wind  load  of  20  Ib.  per  square  foot,  on  a 
vertical  surface;  and  14.9  Ib.  per  square  foot.,  or  1,250  Ib.  per 
panel  acting  normal  to  the  roof  and  on  only  one  side  of  it.  If 
a  wind  load  of  1,250  Ib.  per  panel  is  assumed  to  act  -vertically 
on  both  sides  of  the  truss  in  Fig.  850;  the  resulting  stresses  will 
be  as  given  under  Case  VI  in  Table  IV.  It  is  clear  that  the 
stresses  in  all  chord  members  are  greater  for  Case  VI  than  for 
Cases  I  to  IV  inclusive;  and  that  the  slight  excess  of  the  stresses 
in  the  web  members  for  Cases  I  to  IV  are  negligible  when  the 
additional  stresses  for  dead  and  snow  loads  are  taken  into 
consideration.  In  comparing  Case  V  with  Case  VI,  we  find 
that  the  member  8  requires  the  most  consideration.  Members 
i  and  2  are  the  only  chord  members  which  have  a  stress  greater 
in  Case  V  than  in  Case  VI. 

It  is  clear  that  for  wall  bearing  trusses,  a  load  of  v  =  14.9  Ib. 
per  square  foot  of  roof  surface  taken  vertically  on  both  sides 
of  a  Fink  truss,  having  a  roof  slope  of  6  in.  to  i  ft.,  will  suffice 
for  a  wind  horizontal  load  of  w  =  20  Ib.  per  square  foot  on  a 
vertical  surface.  It  is  also  clear  that  v  varies  directly  as  w. 
Likewise,  if  w  remains  constant  and  the  slope  varies,  the  cor- 
responding value  of  v  may  be  taken  from  Table  II  of  Article 
75.  In  a  wall  bearing  Warren  truss,  having  a  roof  slope  not 
greater  than  i  J^  in.  to  i  ft.,  a  vertical  load  of  5  Ib.  per  square  foot 
seems  to  be  an  ample  allowance  for  wind. 

When  the  truss  is  knee-braced  to  columns,  the  wind  load 


SEC.  II  ROOF    TRUSSES  125 

may  without  serious  error,  be  considered  vertical  and  included 
with  the  dead  and  snow  loads,  if  special  consideration  is  given 
to  the  columns,  knee-braces  and  the  web  members  which 
are  connected  to  the  knee-braces  and  nearest  in  line  with 
them.  It  is  probably  true  that  a  stress  diagram  for  wind  is 
drawn  for  not  more  than  one  truss  for  every  hundred  that  are 
designed. 

79.  Combined   Snow   and   Wind   Loads. — A    considerable 
amount  of  speculation  has  been  made  regarding  the  proper 
combination  of  stresses  caused  by  wind  and  snow.     It  seems 
reasonable  to  predict  that  in  a  given  locality  the  maximum  wind 
and  snow  loads  will  not  occur  simultaneously.     If  the  snow  is 
not  covered  by  a  crust,  a  high  wind  will  blow  it  from  the  roof. 
If  the  roof  is  nearly  flat,  the  snow  load  may  be  large;  but  the 
wind  stresses  will  be  small,  especially  in  a  wall  bearing  truss. 
On  the  other  hand,  the  wind  stresses  are  relatively  larger  on  a 
steep  roof,  but  little  snow  can  be  retained. 

For  buildings  in  the  latitudes  of  New  York  and  Chicago,  it  is 
the  general  practice  to  consider  a  load  of  25  Ib.  per  square  foot 
of  roof  surface  acting  vertically,  as  sufficient  for  combined  wind 
and  snow. 

80.  Design  Stresses. — The  following  data  will  be  assumed: 
Wall  bearing  trusses  (Fig.  830)  15  ft.  apart;  span,  40  ft.,  rise  10 
ft.,   roof   covering  including  purlins  weighs  10  Ib.  per  square 
foot;  combined  snow  and  wind  assumed  at  25  Ib.  per  square 
foot  of  roof  surface.     The  length  of  the  top  chord  is  22.36  ft. 
The  total  weight  on  the  truss  is  35  X  15  X  2  X  22.36  =  23,500 
Ib.,  or  590  Ib.  per  foot  of  span.     From  Table  I  the  assumed 
weight  of  the  truss  is  1,500  Ib.     The  total  load  is  25,000  Ib.  or 
3,125  Ib.  per  panel.     A  stress  diagram  drawn  for  a   vertical 
load  of  3,125   b.,  placed  at  each  of  the  seven  joints  of  the  top 
chord  will  determine  the  design  stresses.     The  half -panel  load 
at  each  support  has  no  influence  upon  the  stresses. 

When  the  same  truss  is  supported  on  columns,  as  in  Fig.  850, 
the  stress  diagram  is  drawn  in  the  same  manner,  using  the 
same  panel  loads  as  for  the  wall-bearing  truss.  All  members 
are  designed  as  for  a  wall-bearing  truss,  except  the  member  A . 
The  size  of  one  angle  necessary  to  carry  the  stress,  as  given  by 


126  THEORY    OF    FRAMED    STRUCTURES  CHAP.  Ill 

the  stress  diagram,  is  determined;  and  two  angles  of  this  size 
are  used. 

The  horizontal  wind  load  on  the  side  of  the  building,  and  the 
horizontal  component  of  the  normal  wind  load  on  the  roof, 
which  are  carried  by  each  bent  are  determined;  and  three- 
eights  of  the  total  is  taken  as  the  horizontal  reaction  of  the 
leeward  column.  In  Article  77  this  reaction  was  found  to  be 
2,920  Ib.  This  reaction,  multiplied  by  two-thirds  the  height 
of  the  column  to  the  foot  of  the  knee-brace,  gives  the  maximum 
bending  moment  at  the  foot  of  the  leeward  knee-brace;  which  is 
2,920  X  10  =  29,200  ft.-lb. 

This  bending  moment  is  obviously  greater  than  at  any  point 
in  the  windward  column.  Each  column  should  be  designed  to 
resist  the  bending  moment  of  29,200  ft.-lb.  and  a  direct  com- 
pression of  12,500  Ib. 

In  designing  the  knee-braces,  the  leeward  knee-brace  should 
be  considered.  The  horizontal  component  of  the  stress,  easily 
determined  from  the  sketch  of  the  leeward  column  in  Fig.  85^;, 
is  7,787  Ib.;  from  which  the  compressive  stress  may  be  quickly 
found. 

81.  Conclusions. — Several  years  ago  a  purchaser  wrote  to  a 
structural  steel  company  for  the  price  of  a  "steel  building,  40 
ft.  wide,  100  ft.  long  and  2  miles  from  the  railroad  station." 
The  author,  who  was  at  the  time  a  designer  for  the  company, 
was  given  the  letter  and  instructed  to  make  a  "design  and  esti- 
mate." While  the  typical  specification  written  by  a  purchas- 
ing agent  gives  more  detail  than  the  one  cited ;  it  is  nevertheless 
true  that  the  cases  are  rare  where  competitive  designs  submitted 
in  accordance  with  the  same  specifications  are  actually  made  on 
the  same  basis.  In  some  designs  no  bending  moment  whatever 
would  be  provided  for  in  the  columns,  in  others  bending  moment 
would  be  considered  on  the  basis  of  16,000  Ib.  per  square  inch 
reduced  for  compression;  while  in  others  24,000  Ib.  per  square 
inch  might  be  used.  But  the  most  interesting  feature  would  be 
found  in  a  comparison  of  the  designs  of  knee-braces  and  bracing 
which  are  generally  a  matter  of  opinion  with  each  designer; 
and  his  opinion  may  vary  with  the  hour  of  the  day  or  the  day  or 
the  week. 


SEC.  II  ROOF    TRUSSES  127 

There  is  no  generally  accepted  standard  specification  for 
buildings  in  America.  Until  one  has  been  written,  the  treat- 
ment of  stresses  as  given  in  Article  80,  though  not  strictly 
scientific,  should  not  be  condemned  by  any  theorist  who  knows 
nothing  of  the  perplexities  of  the  designer  in  playing  the  role 
of  a  mind  reader  while  interpreting  the  average  building 
specifications. 


CHAPTER  IV 
BRIDGES 

SEC.  I.     STANDARD  TYPES 

82.  Standard  Types. — The  most  common  types  of  railway 
steel  bridges  are  shown  in  Figs.  88  to  94.  In  the  deck  plate- 
girder  type  the  cross-ties  which  support  the  rails  rest  directly 
on  the  top  flanges  of  two  girders,  placed  from  6  to  7  ft.  apart 


FIG.   88. — Pratt  Truss. 


FIG.   89. — Warren  Truss. 


FIG.  90. — Parker  Truss. 

In  the  through  plate-girder  type  the  cross-ties  rest  on  short 
deck  beams  or  girders  which  are  supported  at  each  end  by 
cross  girders,  usually  called  floor  beams,  which  are  in  turn  sup- 
ported by  the  main  girders.  The  through  type  is  more  expen- 
sive than  the  deck  type,  and  is  never  used  unless  the  distance 
from  the  rail  to  the  under  side  of  the  main  girders  is  so  small  that 
deck  span  is  impossible.  Plate-girders  are  used  for  spans  up  to 
about  115  ft. 

128 


SEC.  I 


BRIDGES 


I29 


The  Pratt  truss  (Fig.  88)  is  the  common  type  for  pin-con- 
nected trusses  in  which  the  diagonal  web  members  are  eye-bars, 
or  other  slender  members  designed  to  take  tension  only. 

The  Warren  truss  (Fig.  89)  is  built  with  stiff  web  members, 
capable  of  resisting  alternate  stresses  of  tension  and  compression. 
The  Pratt  and  Warren  types  are  used  for  spans  between  about 
115  ft.  and  200  ft. 

The  Parker  truss  (Fig.  90),  more  often  called  a  curved  chord 
truss  or  a  earner-back  truss,  and  often  having  eye-bars  in  the 
bottom  chord  and  diagonal  web  members,  is  the  standard  type 
for  spans  between  about  200  ft.  and  300  ft. 

The  Baltimore  truss  is  shown  in  Figs.  91  and  92.     The  eco- 


FiG.  91. — Baltimore  Truss  (sub-ties). 


PIG.  92. — Baltimore  Truss  (sub-struts). 

nomic  height  of  a  bridge,  being  from  one-fifth  to  one-eighth  of  the 
span,  increases  with  the  length.  The  economic  panel  length 
does  not  increase  in  the  same  ratio,  consequently  as  the  height 
increases  the  diagonals  become  steeper.  From  a  practical  as 
well  as  a  theoretical  standpoint,  the  diagonals  should  not  exceed 
a  slope  of  about  3  vertical  in  2  horizontal.  In  order  to  meet 
this  requirement  in  large  spans,  the  truss  may  be  built  with 
subdivided  panels  as  in  the  Baltimore  truss.  The  floor  beams 
in  the  middle  of  a  long  panel  are  supported  by  a  sub-vertical, 
connecting  to  the  main  diagonal  and  a  sub-tie  as  in  Fig.  92  or  a 
sub-strut  as  in  Fig.  93.  Theoretically  the  Baltimore  truss  has 
no  standing,  for  if  the  length  of  span  makes  sub-paneling  essen- 
tial, a  parallel  chord  truss  is  not  economical. 


13°  THEORY   OF  FRAMED   STRUCTURES  CHAP.  IV 

The  Pennsylvania  truss  with  sub- ties  (Fig.  93)  and  substruts 
(Fig.  94)  is  used  for  all  simple  truss  spans  over  apprximately  300 
ft.  in  length.  The  Metropolis  bridge  over  the  Ohio  River,  720 
ft.  long,  is  the  longest  span  of  this  type  which  has  been  built  to 
date. 

The  standard  types  here  mentioned  are  also  employed  in 
highway  bridges.  Trusses  are  frequently  used  in  highway 
bridges  for  spans  as  short  as  40  or  50  ft. 


FIG.  93. — Pennsylvania  Truss  (sub-ties). 


FIG.  94. — Pennsylvania  truss  (sub-struts). 

83.  Moving  Loads. — The  loads  which  a  roof  truss  supports 
are  usually  considered  stationary;  and  stress  analysis  proceeds 
directly  after  the  determination  of  the  load  and  reactions.     The 
process  is  not  so  simple  in  the  case  of  bridges.     The  loads 
move  upon  the  structure,  and  the  shear  or  bending  moment  at 
any  section,  or  the  stress  in  any  member,  varying  with  the 
movement,  is  required  only  for  that  particular  position  of  the 
loads  which  will  cause  the  maximum  shear,  bending  moment 
or  stress.     The  first  task,  therefore,  is  the  development  of  a 
method  for  finding  the  required  position  of  the  moving  loads 
in  any  given  case.     This  leads  us  to  a  consideration  of  "  In- 
fluence Lines." 

SEC.  II.    DECK  BEAMS  AND  GIRDERS 

84.  Influence  Line  for  Reaction.— Let  P  (Fig.  95)  represent 
a  load  of  i  lb.,  moving  upon  the  beam  AB.     Draw  the  line  QO, 
intersecting  the  verticals  through  A  and  B.    Lay  off  QK  =  i  lb., 
draw  the  line  OK,  and  let  //  represent  the  ordinate  under 
the  load  P.     As  P  moves  from  B  to  A,  the  reaction  at  A  in- 


SEC.  II 


BRIDGES 


creases  uniformly  from  zero,  when  the  load  is  at  B;  to  i  lb., 
when  the  load  reaches  A .  Likewise  the  ordinate  //  increases 
uniformly  from  zero  at  O  to  QK  =  i  lb.,  at  Q.  Hence  for  any 
position  of  the  load,  the  ordinate  //,  directly  under  the  load, 
represents  the  reaction  at  A .  The  line  OK  is  called  an  influence 
line  for  the  reaction  at  A ,  since  the  ordinate  //  shows  how  the 
reaction  at  A  varies  or  is  influenced  by  any  movement  of  the 
load. 

The  influence  line,  thus  drawn  by  the  aid  of  the  unit  load, 
may  be  used  for  finding  the  reaction  at  A  for  a  series  of  loads 
in  any  position  by  multiplying  the  ordinate  under  each  load 
by  the  ratio  of  its  weight  to  i  lb.,  and  adding  the  products. 
In  Fig.  96  the  ordinates  for  the  loads  Q  and  S  are  0.7  and  0.4 


o 


Q-IOOO* 

o 


3000* 


O 


5=2000* 


^' 


B 


FIG.  95-  FIG.  96. 

respectively.     Hence,  the  reaction  at  A  is 

2,000  X  0.4  =  800 
1,000  X  0.7  =  700 

1,500  lb. 

Since  the  influence  line  has  a  constant  slope,  the  reaction  at  A 
may  also  be  determined  by  finding  the  product  of  the  total 
load  (3,000),  and  the  ordinate  at  the  center  of  gravity  of  the 
loads;  thus, 

3,000  X  0.5  =  1,500  lb. 

The  load  of  i  lb.,  which  is  invariably  used  in  constructing 
influence  lines,  may  be  considered  as  acting  the  role  of  a  scout, 
sent  across  the  span  in  advance  of  other  loads.  The  effect 
of  this  advance  load  upon  the  reaction,  when  represented 
graphically,  becomes  an  influence  line  for  the  reaction.  The 
same  is  true  in  the  case  of  shear  or  bending  movement.  Influ- 


132  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

ence  lines  thus  drawn  may  be  used  as  a  basis  for  determining 
the  effect  produced  by  a  series  of  loads;  such  as  a  road  roller, 
traction  engine,  or  a  railway  train. 

85.  Influence  Line  for  Shear  in  a  Beam. — Let  us  first  have  a 
clear  understanding  of  what  we  mean  by  shear. 

The  shear  at  any  normal  section  of  a  beam  is  the  algebraic 
sum  of  all  the  transverse  forces  acting  on  one  side  (either  side) 
of  the  section.  When  this  sum  or  resultant  force  acts  upward 
on  the  left  of  the  section,  the  shear  is  positive.  Obviously, 
if  the  resultant  of  the  forces  on  the  left  of  the  section  acts 
upward,  the  resultant  of  the  forces  on  the  right  of  the  section 
acts  downward. 

The  influence  line  for  shear  at  section  C  (Fig.  97)  is  desired. 

When  the  load  of  i  Ib.  is  between 
B  and  C,  the  only  force  acting  on 
the  left  of  the  section  is  the  reac- 
tion at  A.  Hence,  the  influence 
line  OK  for  the  left  reaction,  is 
also  the  influence  line  for  shear  at 
FlG  9?  C,  when  the  load  is  between  B 

and  C. 

When  the  load  of  i  Ib.  is  between  A  and  C,  the  only  force  on 
the  right  of  the  section  is  the  right  reaction  acting  upward;  and 
the  shear  at  the  section  is  the  right  reaction  taken  negatively. 
Hence,  the  line  QH,  being  the  influence  line  for  the  right  re- 
action taken  negatively,  is  the  influence  line  for  shear  at  C 
when  the  load  of  i  Ib.  is  between  A  and  C.  Therefore  the  line 
QSNTO  is  the  influence  line  for  shear  on  the  section  at  C;  and 
shows  that  when  a  single  load  (whether  i  Ib.  or  20  tons)  crosses 
the  span  from  B  to  A ,  the  shear  at  C  increases  uniformly  until 
the  load  is  just  at  the  right  of  C.  When  the  load  crosses  the 
point  C,  the  shear  at  C  instantly  becomes  negative,  increasing 
to  zero  when  the  load  arrives  at  A . 

86.  Influence  Line  for  Bending  Moment  in  a  Beam. — The 
bending  moment  at  any  normal  section  of  a  beam  is  the  alge- 
braic sum  of  the  moments  of  all  the  forces  acting  on  one  side 
(either  side)  of  the  section,  taken  about  the  center  of  gravity 
of  the  section  as  an  axis.     When  this  sum  or  resulting  moment 


SEC.  H 


BRIDGES 


133 


«..  ../o '---->< 75 -> 


is  clockwise  on  the  left  of  and  about  the  section,  the  bending 
moment  is  positive.  Obviously  if  the  sum  of  the  moments  on 
the  left  of  the  section  is  clockwise,  the  sum  of  the  moments  on 
the  right  of  the  section  is  counter-clockwise. 

The  influence  line  for  the  bending  moment  at  section  C  (Fig. 
98)  is  desired.  When  the  load  of  i  Ib.  is  between  B  and  C,  the 
only  force  acting  on  the  left  of  the  section  is  the  left  reaction  at 
A ;  and  the  bending  moment  at  C  equals  10  ft.  times  the  left 
reaction.  As  the  load  moves  from  B  to  A,  the  left  reaction 
increases  uniformly  from  o  to  i  Ib.,  and  10  ft.  times  the  left 
reaction  increases  uniformly  from  o  to  10  ft.-lb.  Likewise  the 
ordinate  to  the  line  OK}  increasing  uniformly  from  zero  at  O  to 
10  ft.-lb.  at  Q,  represents  10  ft.  times  the  left  reaction.  Hence 
the  line  ON  is  the  influence  line 
for  the  bending  moment  at  C 
when  the  load  of  i  Ib.  is  between 
B  and  C.  The  ordinates  are 
positive,  since  the  moment  of 
the  force  on  the  left  of  the  sec- 
tion is  clockwise. 

When  the  load  of  i  Ib.  is  be- 
tween C  and  A,  the  only  force 
acting  on  the  right  of  the  section  is  the  right  reaction  at  B ;  and 
the  moment  at  C  equals  15  ft.  times  the  right  reaction.  As  the 
load  moves  from  A  to  C,  the  right  reaction  increases  uniformly 
from  o  to  i  Ib.;  and  15  ft.  times  the  right  reaction  increases 
uniformly  from  o  to  15  ft.-lb.  Likewise  the  ordinate  of  the 
line  QHj  increasing  uniformly  from  zero  at  Q  to  15  ft.-lb.  at  0, 
represents  15  ft.  times  the  right  reaction.  Hence,  the  line  QN 
is  the  influence  line  for  the  bending  moment  at  C,  when  the  load 
of  i  Ib.  is  between  A  and  C.  These  ordinates  are  also  positive, 
since  the  moment  of  the  force  on  the  right  of  the  section  is 
counter-clockwise.  Therefore,  the  line  QNO  is  the  influence 
line  for  the  bending  moment  at  C;  and  shows  that  when  a 
single  load  (whether  i  Ib.  or  100  Ib.)  crosses  the  span  from  B 
to  A ,  the  bending  moment  at  C  increases  uniformly  to  a  maxi- 
mum as  the  load  arrives  at  C,  and  decreases  uniformly  to  zero 
as  the  load  arrives  at  A . 


15* 


FIG.  98. 


134 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


87.  Criterion  for  Maximum  Bending  Moment.  —  Consider  a 
series  of  loads  so  connected  that  the  distances  between  loads 

are  maintained  (Fig.  99)  .  As  these 
loads  move  on  the  span,  the  bend- 
ing moment  at  C  varies.  It  is  de- 
sired to  find  the  position  of  these 
loads  on  the  span,  so  that  the  bend- 
ing moment  at  C  will  be  a  maxi- 
mum. The  influence  line  QNO  is 
drawn  in  accordance  with  the 
preceding  article.  The  bending 
moment  at  C,  for  any  given  posi- 
tion of  the  loads,  may  be  deter- 
mined by  multiplying  each  load 
by  its  corresponding  ordinate  in  the  influence  line  diagram; 
and  finding  the  sum  of  these  products.  Or,  since  the  influence 
line  has  a  constant  slope  from  Q  to  N,  and  a  constant  slope  from 
N  to  O;  the  bending  moment  at  C  may  be  expressed  by  the 
equation 

Mc  =  P&z  +  Piyi 

in  which  PI  =  the  sum  of  the  loads  between  A  and  C 
PZ  =  the  sum  of  the  loads  between  C  and  B 
yi  =  the  ordinate  at  the  resultant  PI 

and  yz  =  the  ordinate  at  the  resultant  PZ- 

Let  the  loads  be  moved  any  small  distance  d  to  the  left;  in 
such  a  way  that  no  loads  pass  the  points  A,  C  and  B.  In  other 
words  PI  and  PZ  remain  constant  while  the  movement  takes 
place.  The  bending  moment  at  C  for  this  new  position  is 

M'c  =  P^(yz  +  yzz)  +  Pi(yi 
The  change  in  the  bending  moment  at  C  is 

AAf  C   =    M'c   —   M  C   =   PzyZZ 

Has  the  bending  moment  at  C  been  increased  or  diminished  by 
this  movement?  The  answer  to  this  question  depends  upon 
whether 


As  long  as 


<Pzyzz 


SEC.  II  BRIDGES  135 

the  bending  moment  at  C  is  increased  by  the  movement  of  the 
loads  to  the  left. 

From  similar  triangles 
and  yn:<::7o:ioo 


, 
or  3/11  =  -  —  and  y22  = 

100  100 

Hence  the  bending  moment  at  C  will  increase  as  long  as 


IOO  IOO 

or  as  long  as  7Pi<3P2. 

Let  PI  +  P2  =  P  =  total  load  on  the  span. 

If  7^i  <     3^2 

and  3Pi  =  3Pi 

then  ioPi<3P 

or  Pi<  —  P. 

10 

Hence,  the  bending  moment  at  C  will  increase  by  a  movement 
of  the  loads  to  the  left  as  long  as 


10 

If  no  additional  loads  move  on  to  the  span  at  B  as  the  loads 
move  to  the  left,  and  none  pass  off  at  A  ,  the  total  load  P  on  the 
span  remains  constant;  but  the  loads  PI  on  the  segment  AC 
are  increased  whenever  a  load  passes  the  point  C. 

Now  there  will  be  a  certain  load  which,  when  approaching  C, 
will  render 

Pi<^P, 

10 

and  the  bending  moment  at  C  will  be  increasing;  but,  having 
passed  C  and  become  a  part  of  PI,  will  render 

Pi>^P, 

10 

in  which  case  the  bending  moment  at  C  will  be  decreasing. 
This  particular  load  is  called  the  critical  load,  for  when  it  is  at  C, 
the  bending  moment  at  C  is  a  maximum. 


i36 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VI 


MOMENT  TABLE 

2-142  Ton  Engines  followed  by  4000  Ib. 
COOPER'S  CLASS  E-40 


per  ft. 


3 


3- 


3- 


88 


FIG.  100. 


SEC.  II  BRIDGES  137 

Thus  we  see  how  an  influence-line  diagram  may  be  used  to 
develop  the  criterion  for  maximum  bending  moment  at  any 
section  of  a  beam  of  any  length.  After  the  correct  position 
of  the  loads  has  been  determined,  the  maximum  bending 
moment  may  be  computed  by  scaling  the  length  of  the  ordinate 
under  each  load;  and  taking  the  sum  of  the  products  of  each 
load  and  its  corresponding  ordinate. 

88.  Cooper's  Standard  Train  Loads. — When  the  series  of 
loads  represents  the  wheel  concentrations  of  a  railway  train, 
the  bending  moment  may  be  more  easily  computed  by  the 
help  of  a  "Momemt  Table";  as  shown  in  Fig.  100.  This 
table  is  a  collection  of  data  for  what  is  known  as  a  "Cooper's 
£-40  Loading" ;  consisting  of  two  142-ton  locomotives,  followed 
by  a  uniform  train  load  of  4,000.  Ib.  per  linear  foot.  Each  axle 
for  the  large  wheel  supports  a  load  of  40,000  Ib.  A  train  is 
usually  supported  by  two  girders  or  trusses,  one-half  of  the  load 
coming  upon  each.  For  this  reason,  wheel  loads  or  the  loads 
supported  by  each  half  of  the  structure,  instead  of  axle  loads, 
are  given  in  the  table.  The  wheel  loads  are  given  on  line  c; 
wheel  i  supports  10,000  Ib.;  wheel,  4,  20,000  Ib.;  wheel  16, 
13,000  Ib.;  etc.  Accumulative  loads  are  given  on  lines  a  and  b. 
The  spacing  of  the  wheels  is  given  on  line  d,  and  the  accumula- 
tive distances  on  lines  e  and/.  The  quantities  given  below  line 
/  are  moments  expressed  in  units  of  1,000  ft.-lb. 

Suppose  the  moment  is  required  of  all  the  wheels  to  the  left 
of,  and  about  wheel  6.  Follow  the  space  between  wheels  5  and 
6  down  to  the  heavy  line;  and  then  to  the  left,  under  wheel  i, 
read  1,640.  The  moment  of  the  wheels  up  to,  and  including 
wheel  3,  about  wheel  6  is  840.  Verify  these  quantities  by 
computation.  The  quantities  below  the  heavy  zigzag  line 
are  moments  on  the  right  of  various  wheels,  instead  of  on  the 
left. 

Cooper's  Loadings  are  specified  more  extensively  in  American 
railway  practice  than  in  any  other.  They  are  divided  into 
classes  and  are  specified  as  Cooper's  £-40,  £-50,  £-55,  etc. 
The  wheel  spacing  in  all  classes  is  the  same.  A  constant  ratio 
exists  between  the  loads  of  one  class  and  the  corresponding 
loads  of  any  other  class.  Thus  the  loads  of  a  Cooper's  E-6o 


138  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

are  50  per  cent  greater  than  those  for  an  £-40,  given  in  Fig.  100. 
If  an  E-6o  is  specified,  the  table  of  Fig.  100  may  still  be  used, 
if  the  final  results  are  increased  by  50  per  cent. 
89.  Illustrative  Problem. 

Determine  the  maximum  bending  moment  for  an  £-40  load- 
ing at  the  center  of  a  62-ft.  span.     Draw  the  influence  line  dia- 

y   gram  and  develop   the  criterion 

A         n    OO@Q  —  o  o  pom     for  maximum   bending  at  point 

kv-3/'  .....  -^----^----J    -  c  (Fig.  101).     When  developed, 

FlG'  I01'  this  criterion  will  show  that,  as 

the  train  comes  on  the  span  at  B,  the  bending  moment  at  C 

will  increase  as  long  as  the  loads  on  AC  are  less  than  one-half  the 

total  load  on  the  span.     The  influence  line  shows  that  as  many 

loads  as  possible  should  be  placed  on  the  span,  with  the  heavy 

loads  usually  (not  always)  near  C,  where  the  ordinates  are  long. 

Try  wheel  4  at  C.     Wheel  4  is  18  ft.  from  the  head  of  the 

train,  hence  the  length  of  train  on  the  span  is  18  +  31  =  49  ft. 

Wheel  9,  being  48  ft.  from  the  head  of  the  train,  is  i  ft.  on  the 

span. 

2)142  =  total  load  on  the  span 

71  =  one-half  the  load  on  the  span 

50  =  load  on  AC  when  wheel  4  is  approaching  C 

70  =  load  on  AC  when  wheel  4  has  passed  C. 


In  either  case  the  load  on  A  C  is  less  than  one-half  the  load  on 
the  span,  and  the  bending  moment  at  C  continues  to  increase 
as  wheel  4  crosses  C. 
Try  wheel  5  at  C. 

Make  a  sketch  similar  to  Fig.  101  for  this,  and  each  succeeding 
case.  In  this  arrangement  the  large  wheels  are  not  so  well 
located  about  C;  but  the  loads  as  a  whole  are  nearer  the  center 
of  the  span. 

2)142 

7  1  >  70  when  wheel  5  is  approaching  C 

<9<D  when  wheel  5  has  passed  C. 
When  wheel  5  is  approaching  C,  the  load  on  A  C  is  less  than  one- 


SEC.  II  BRIDGES  139 

half  the  load  on  the  span,  and  the  bending  moment  at  C  is 
consequently  increasing;  but  when  wheel  5  has  passed  C,  the 
load  on  A  C  is  greater  than  one-half  the  load  on  the  span,  and  the 
bending  moment  at  C  is  consequently  decreasing.  Hence 
wheel  5  at  C  satisfies  the  criterion  for  maximum  bending 
moment  at  C. 

Try  wheel  6  at  C. 

Wheel  i  has  now  moved  the  off  span,  and  wheel  10  has  come  on ; 
causing  no  change  in  the  total  load  on  the  span. 

2)142 

71  <8o  bending  moment  at  C  is  decreasing,  with  wheel  6 

approaching  C. 

<93  bending  moment  at  C  is  decreasing,  with  wheel  6 
passing  C. 

It  is  to  be  noted  that,  as  the  loads  move  to  the  left  on  the  span, 
the  bending  moment  at  C  increases  from  zero,  when  wheel  i 
is  at  B\  to  a  maximum,  when  wheel  5  is  at  C.  As  wheel  5 
passes  C,  the  bending  moment  at  C  begins  to  decrease.  To  show 
this,  the  bending  moment  at  C  will  now  be  computed  for  the 
three  positions  of  the  load  when  wheels  4,  5  and  6  are  at  C. 

Wheel  4  at  C. 

We  shall  compute  the  bending  moment  at  C  by  taking  the 
algebraic  sum  of  the  moments  of  all  the  forces  acting  on  the 
left  of  C.  These  forces  are  the  reaction  at  A  and  wheels  i,  2 
and  3.  The  work  is  greatly  simplified  by  using  the  moment 
table.  We  shall  first  find  the  moment  of  all  the  wheels  about 
B.  The  moment  of  all  the  wheels  to  the  left  of,  and  about  9, 
is  3,496  as  given  in  the  table.  The  moment  of  all  the  wheels 
about  B  equals  the  moment  of  all  the  wheels  about  wheel  9; 
plus  the  weight  of  all  the  wheels  times  i  ft. 

3,496 
142  X  i  =         142 

3,638  =  the  moment  of  all  the  loads  on  the 

span  about  B. 
The  reaction  at  A  is 


140  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

The  moment  of  all  the  wheels  to  the  left  of,  and  about  C  is  480, 
hence  the  bending  moment  at  C  is 


480 


1,339  =  bending  moment  at  C. 
Wheel  5  at  C. 

3.496 
142  X    6  =     852 

4,348  =  moment  about  B. 

-Lj—  X  31  =  2,174 
830 
1,344  =  bending  moment  at  C. 

Wheel  6  at  C. 

4,072  =  moment  about  wheel  10  of  wheels  up  to 
142  X  7  =       994  and  including  wheel  2 

2)5,066 

2,533  =  moment  of  the  left  reaction  about  C. 
1,320  =  moment  about  wheel  6  of  wheels  up  to 

and  including  wheel  2. 
1,213  ~  bending  moment  at  C. 
The  results  are  tabulated  below: 

AT  C  BENDING  MOMENT 

Wheel  4 I>339  increasing 

Wheel  5 i->344  a  maximum 

Wheel  6 1,213  decreasing 

Let  us  continue  the  investigation  further. 
Try  wheel  7  at  C. 

2)162 

8 1  <  93  decreasing 
<io6  decreasing 

Try  wheel  8  at  C. 
Note  that  when  wheel  8  is  approaching  C,  wheel  13  is  not  on  the 


SEC.  II  BRIDGES  141 

span  and  the  total  load  is  162;  but  when  wheel  8  has  passed  C, 
wheel  13  is  on  the  span,  and  the  total  load  is  182;  hence, 

2)162 

8i<  86  decreasing 
2)182 

91  <   99  decreasing. 

Try  wheel  9  at  C. 

2)162 

8i>  79  increasing 
2)182  \  a  maximum 


91  <92  decreasing 

Since  the  bending  moment  was  decreasing  when  wheel  8 
had  passed  C,  and  was  increasing  when  wheel  9  was  arriving 
at  C;  there  must  have  been  some  intermediate  position,  at  which 
the  bending  moment  was  a  minimum.  This  position  occurs 
when  wheel  3  leaves  the  span  at  A  . 

Wheel  3  at  A. 
2)182 

91  <99,  wheel  3  approaching  A,  decreasing  j 
2)162  >  a  minimum 

8i>  79,  wheel  3  passing  A,  increasing 
Wheel  10  at  C. 
2)142 

7  1  >  5  2  1  increasing 

>62    J 

Wheel  4  at  A 
2)182 

91  <92  decreasing  1 
2)162^  >  a  minimum. 

8i>72  increasing 
Wheel  ii  at  C. 


77.5>49  \  . 

>69  /^creasing 


142  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

Wheel  12  atC. 


77-5>56 

J     °    increasing 

Wheel  13  at  C. 
2)168 

84  >  76  increasing 
2)155  a  maximum. 

7  7.  5  <  83  decreasing 

Wheel  14  at  C. 


78  5  <  83  decreasing 
2)144 

72  <go  decreasing. 

Check  the  bending  moments  at  C  for  the  several  positions  of 
the  train,  given  in  the  following  table  : 

BENDING  MOMENT  IN  UNITS 
AT  C  AT  A  OF  1000  FT.-LB. 


Wheel  4 

5 
6 

7 
8 


10 
ii 


12 


Wheel  3 
Wheel  4 


,339 

,344  a  maximum 

,213 

,143 

,083 

,075  a  minimum 

,083  a  maximum 

,082  a  minimum 

,165 

,302 

,357-5 


13  i ,  3  7 1 . 5  the  maximum 

14  1,344-5 

Wheels  5,  9  and  13  at  C  satisfy  the  criterion  for  a  maximum 
bending  moment  at  C;  and  wheels  3  and  4  at  A  satisfy  the 
criterion  for  a  minimum  bending  moment  at  C.  It  frequently 
happens,  as  in  the  present  case,  that  more  than  one  position  of 
the  train  will  satisfy  the  criterion  for  maximum  bending  mo- 
ment at  a  given  section.  For  each  two  consecutive  positions, 
causing  a  maximum  bending  moment;  there  necessarily  exists 


SEC.  II  BRIDGES  143 

an  intervening  position,  causing  a  minimum  bending  moment. 
Only  the  maximum  values  have  a  practical  significance.  When 
several  positions  of  the  train  satisfy  the  criterion  for  a  maximum 
bending  moment,  it  becomes  necessary  to  compute  the  bending 
moment  for  each  position  to  determine  the  greatest  maximum 
value. 

90.  Problems. 

Compute  the  maximum  bending  moment  at  the  center  of  a 
girder  having  a  span  of  100  ft.,  for  an  £-40  loading. 

Draw  the  influence  line,  develop  the  criterion  and  compute 
the  maximum  bending  moment  at  four  sections,  10  ft.  apart, 
between  the  ends  and  the  center.  Remember  that  the  train 
may  come  on  to  the  span  from  either  end. 

9  1  .  General  Criterion  for  Maximum  B  ending  M  om  ent.  —  In 
Fig.  99  let  AC  =  a  and  AB  =  I.  Draw  the  influence  line  for 
bending  moment  at  C;  in  which  QK  =  a  ft.-lb.,  and  OH  =  l—a 
ft.-lb.  Develop  the  criterion,  showing  that  for  maximum 
bending  moment  at  C, 

P^>jP  tt 

This  criterion  is  often  expressed  in  the  form 


(2) 


-.-Fr  (3) 

a        I        I  —  a 

and  may  be  stated  as  follows:  A  maximum  bending  moment 
at  any  point  occurs  whenever  the  train  is  so  placed  that  the 
load  on  each  segment,  divided  by  the  length  of  the  segment, 
equals  the  total  load  on  the  span,  divided  by  the  span  length. 
In  other  words,  the  average  load  per  foot  on  each  segment, 
equals  the  average  load  per  foot  on  the  span.  In  interpreting 
Eqs.  (2)  and  (3),  PI  represents  the  load  on  the  segment  a  plus  a 
portion  of  the  load  at  C;  the  remainder  of  the  load  at  C  being 
considered  as  a  part  of  P2.  If  Eqs.  (2)  and  (3)  are  satisfied 
when  no  load  is  at  C,  any  movement  of  the  train  will  cause  no 
change  in  the  bending  moment  at  C  as  long  as  PI  and  PI 
remain  constant. 

92.  The  Point  of  Greatest  Maximum  Bending  Moment  in  a 
Beam.—  By  comparing  the  bending  moments  at  the  different 


144 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.   IV 


points  of  the  loo-ft.  girder  in  the  problem  of  Article  90,  the 
student  may  conclude  that  the  maximum  bending  moment  at 
the  center  of  any  beam  is  greater  than  at  any  other  point. 
Such  a  conclusion  is  usually  erroneous.  The  maximum  bending 
moment  at  the  center  of  the  62-ft.  span,  previously  considered, 
was  1,371.5  ft.-lb.,  when  wheel  13  was  at  the  center  and  wheel 
18  was  i  ft.  on  the  span.  When  wheel  13  is  32.3742  ft.  from 
the  right  end  of  the  span  (Fig.  102),  the  bending  moment  under 
wheel  13  is  1376.2  ft.-lb.,  which  is  greater  than  the  bending 
moment  at  the  center. 

Let  X  (Fig.  103)  represent  the  point  of  greatest  maximum 
bending  moment  in  a  beam.     If  the  influence  line  for  bending 


o  O    O  o  O  o  o    n 


1 

1  *.--  -29.6258=  b--> 
\<—  -27.2516  '--->! 

o    o    OCX 

5>O     n  o  o  i 

t 

-29.6258=a--> 

\<  --32.3742  '  —  > 
FIG.  102. 

< -23742 


t-a 


FIG.  103. 

moment  at  X  is  drawn,  and  the  criterion  for  maximum  bending 
moment  developed,  we  shall  have 


a  P 


as  the  wheel  at  X  is  considered  on  the  right  or  the  left  of  X . 

Also,  if  the  bending  moment  is  a  maximum  at  X,  the  shear 
changes  from  positive  to  negative  at  X.  Let  R  represent  the 
left  reaction,  then 

as  the  wheel  at  X  is  considered  on  the  right  or  the  left  of  X. 
Let  b  represent  the  distance  from  B  to  the  center  of  gravity 
of  all  the  loads  on  the  span;  then 

*-r?        -    , 


therefore 


SEC.  II  BRIDGES  145 

a      <  b 
whence  7  P  >  j  P 

or  a  >  b 

as  the  wheel  at  X  is  considered  on  one  side  or  the  other  of  X. 

Obviously,  when  the  wheel  is  at  X 

a  =  b 

When  wheel  13  is  at  the  center  of  the  span  (Fig.  102),  wheel  8 
is  over  the  left  support,  and  consequently  not  on  the  span. 
155)4,224 

27.2516 

The  center  of  gravity  of  wheels  9  to  18  inclusive  is  27.2516  ft. 
from  wheel  18;  or  2.7484  ft.  to  the  right  of  wheel  13. 

62. 

2.7484 

2)59.2516  =  a  +  b 
29.6258  =  a  =  b 

The  center  of  gravity  of  the  total  load  is  29.6258  ft.  from  the 
right  end  of  the  span;  and  the  point  of  maximum  bending 
moment  is  under  wheel  13,  which  is  29.6258  ft.  from  the  left  end 
of  the  span.  The  maximum  bending  moment  is  1376.2  ft.-lb. 

The  point  of  maximum  bending  moment  is  at  the  center  of 
the  span  only  when  the  center  of  gravity  of  the  wheels  on  the 
span  coincides  with  the  wheel  which  causes  the  maximum 
bending  moment  at  the  center.  Otherwise  the  point  of  maxi- 
mum bending  moment  is  not  at  the  center;  but  near  the  center, 
under  the  wheel  which  causes  maximum  bending  moment  at 
the  center;  when  that  wheel  is  as  far  from  one  end  of  the  span 
as  the  center  of  gravity  of  the  loads  on  the  span  is  from  the 
other  end. 

93.  Problems. 

A  road  roller  is  supported  on  two  axles,  9  ft.  apart.  The  load 
on  the  front  axle  is  10,000  Ib.  The  load  on  the  rear  axle  is 
20,000  Ib.  Compute  the  maximum  bending  moment  (a)  in  a 
span  21  ft.  long:  (b)  in  a  span  13  ft.  long.  What  is  the  greatest 
maximum  bending  moment  in  a  loo-ft.  span  for  an  £-40  train? 
For  an  E-6o  train  ? 


146  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

94.  Criterion  for  Maximum  Shear  in  a  Beam. — By  referring 
to  the  influence  line  for  shear  (Fig.  97),  it  is  apparent  that  the 
positive  shear  at  C  increases  from  zero  to  a  maximum,  as  wheel 
i  comes  on  the  span  at  B  and  moves  to  C.     As  wheel  i  crosses 
the  point  C,  the  shear  at  C  is  instantly  diminished  by   the 
amount  of  wheel  i.     Evidently  a  maximum  shear   occurs  as 
each  successive  wheel  arrives  at  C,  followed  by  a  minimum 
shear,  as  the  wheel  crosses  C.     When  wheel  i  is  at  the  center 
of  a  loo-ft.  span,  the  shear  at  the  center  equals  the  left  reaction, 
or  37.8.     This  shear  is  instantly  decreased  to  27.8,  as  wheel  i 
crosses  the  center.     When  wheel  2  comes  up  to  the  center,  the 
shear  is  49.36  —  10  =  39.36;  instantly  decreasing  to  19.36  as 
wheel  2  crosses  the  center;  and  increasing  to  26.96  as  wheel  3 
arrives  at  the  center.     In  the  ordinary  beam  or  deck  plate- 
girder  span,  for  which  the  influence  for  shear  in  Fig.  97  is 
typical,  the  maximum  positive  shear  on  any  section  in  the  left- 
half  of  the  span  will  usually  (not  always)  occur  when  wheel  2 
is  at  the  section. 

95.  Problem. 

Compute  the  maximum  positive  shear  for  an  £-40  loading 
for  sections  10  ft.  apart,  from  the  center  to  the  left  end  of  a 
loo-ft.  deck  plate-girder  span. 

SEC.  III.     PARALLEL  CHORD  TRUSSES 

96.  A  railway  truss  bridge  is  shown  in  Fig.  104,  to  illustrate 
the  manner  in  which  the  wheel  loads  of  a  train  are  carried  to 
and  supported  by  the  piers  at  either  end.     The  wheels  rest 
upon  the  rails  which  are  supported  by  cross- ties.     The  ties 
rest  on  longitudinal  stringers  which  are  supported  by  cross- 
beams, more  often  called  floor  beams.     The  floor  beams  are 
supported  by  the  truss  at  the  bottom  chord  panel  points.     It 
should  be  thoroughly  understood,  and  constantly  kept  in  mind, 
that  the  truss  does  not  directly  support  any  wheel  loads.     The 
truss  receives  the  weight  of  the  train  only  in  the  form  of  floor  beam 
loads,  delivered  to  the  truss  at  the  panel  points. 

97.  Influence  Line  for  Maximum  Stress  in  a  Chord  Member. 
Let  the  truss  in  Fig.  105  be  loaded  in  any  manner,  and  let 
Mz   represent  the  algebraic  sum   of  the  moments  about  Lz 


SEC.  Ill 


BRIDGES 


147 


of  all  the  external  forces  acting  on  one  side  (either  side)  of  the 
section  through  the  panel  1-2.  Since  the  stress  in  UiUz  varies 
as  Mz,  it  is  clear  that  the  stress  in  UiUz  is  a  maximum,  when 


U, 


501 


\ 


LO      L, 


i 


PIG.  104. 

Mi  is  a  maximum.     For  this  reason  we  shall  draw  the  influence 
line  for  Mi  as  a  load  of  i  Ib.  moves  across  the  span. 

When  the  load  of  i  Ib.  is  between  L4  and  Lz,  the  only  force 
acting  on  the  left  of  the  section  is  the  left  reaction  R0,  and  Mz 
=  $oR0.  As  the  load  moves  from 
L±  to  Lo,  RQ  increases  uniformly 
from  o  to  i  Ib.,  and  5cxRo  increases 
uniformly  from  o  to  50  f t.-lb.  Hence 
the  line  ON  is  the  influence  line  for 
Mz  when  the  load  of  i  Ib.  is  between 
L±  and  Lz. 

When  the  load  of  i  Ib.  is  between 
Lo  and  Li,  the  only  force  acting  on 
the  right  of  the  section  is  the  right  reaction  Rt,  and  Mz  = 
5<xR4.  When  the  load  is  between  LI  and  L%,  there  are  two 
forces  acting  on  the  right  of  the  section — a  floor  beam  load  at 
Lz  and  the  truss  reaction  R±',  and  the  moment  of  these  two 
forces  about  L%  is  also  MI  =  50^4.  As  the  load  moves  from 
Lo  to  L4,  Rt  increases  uniformly  from  o  to  i  Ib.;  and  50^4 
increases  uniformly  from  o  to  50  ft.-lb.  Hence,  the  line  QN 


FIG.  105. 


148 


THEORY    OF    FRAMED    STRUCTURES' 


CHAP.  IV 


is  the  influence  line  for  Mz,  when  the  load  of  i  Ib.  is  between 
LQ  and  Z,2. 

Now  it  may  be  observed  that  the  influence  line  ONQ  for  Mz 
is  identical  with  the  influence  line  which  was  drawn,  when  the 
criterion  for  maximum  bending  moment  at  the  center  of  the 
ico-ft.  girder  was  required  in  Article  90.  Hence,  the  criterion 
and  position  of  an  £-40  train  are  the  same  in  each  case.  It  Was 
found  in  Article  90  that  wheels  u,  12  and  13  at  the  center, 


PIG.  i 06. 

satisfied  the  criterion  for  a  maximum  bending  moment;  wheel 
12  giving  the  greatest  maximum,  or  3,219  ft.-lb. 

98.  Illustrative  Problem. 

Determine  the  stress  in  U\Uz  for  an  £-40  loading  when  wheel 
1 2  is  at  Lz  (Fig.  106). 

There  are  seven  forces  acting  on  the  truss — five  floor  beam 
loads  and  two  truss  reactions.  After  the  floor  beam  loads 
and  the  resulting  truss  reactions  have  been  determined, 
the  stress  in  UiUz  may  be  obtained  by  finding  the  sum  of 
the  moments  about  LZ  of  all  the  forces  acting  on  one  side 
(either  side)  of  a  section  through  panel  1-2,  and  dividing  the 
sum  by  32. 

of  the  floor  beam  loads  about  Li. 
23.80  X      o  =         o 
56.56  X     25  =  1,414 

74.52  x  50  =  3.726 

51.92  X    75  =  3,894 
27.20  X  ioo  =  2,720 
234.00          100)11,754 

117.54  =  truss  reaction  at  LQ 


SEC.  Ill  BRIDGES  149 

SM  about  LI  of  the  forces  on  the  left  of  the  section. 

117.  54  X  50    =  5,877 

51.92  X  25    =  1,298 
27.20  X  50    =  1,360  2,658 

32)3,  2I9 

100.6  =  compressive  stress  in  UiUz- 

The  problem  may  be  solved  more  quickly  by  considering  the 
wheel  loads  instead  of  the  floor  beam  loads,  and  using  the 
moment  table. 

I.M  about  L4. 


214  X  10  =  2,140 

I02   =         TOO 

TOO)  11,754 

117-54  truss  reaction  at  L0. 

117.54  X  50  =  5,877  =  moment  about  L2  of  the  left  reaction 
2,658  =  moment  about  wheel  12  of  all  wheels 
32)3,219  from  5  to  12  inclusive. 

100.6  =  compressive  stress  in  UiUz- 

99.  General  Criterion  for  Maximum  Chord  Stress.  —  Let 

Mi  represent  the  algebraic  sum  of  the  moments  about  17%  or  L2 
of  all  the  external  forces  acting  on  one  side  (either  side)  of  the 
section  through  panel  1-2  of  either  truss  (Fig.  107),  as  a  load 
of  i  Ib.  moves  across  either  span.  When  the  load  of  i  Ib.  is 
between  L6  and  L2,  the  only  force  acting  on  the  left  of  the  section 
is  the  left  reaction  RQ,  and  Mz  =  aRo-  As  the  load  moves  from 
L6  to  L0,  RO  increases  uniformly  from  o  to  i  Ib.,  and  Mz  =  aRo 
increases  uniformly  from  o  to  a  ft.-lb.  Likewise  the  ordinate 
of  the  line  OK,  increasing  uniformly  from  o  at  L6  to  a  ft.-lb.  at 
Lo,  represents  aR0.  Hence,  the  line  ON  is  the  influence  line 
for  Mz,  when  the  load  of  i  Ib.  is  between  L6  and  L2. 

When  the  load  of  i  Ib.  is  between  L0  and  LI,  the  only  force 
acting  on  the  right  of  the  section  is  the  right  reaction  R6. 
When  the  load  is  between  LI  and  L2,  there  are  two  forces  acting 
on  the  right  of  the  section  —  a  floor  beam  load  at  L2  and  the 
reaction  R6.  Whether  the  load  of  i  Ib.  is  in  the  first  or  second 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


panel,  M 2  =  (/  —  a)R6.  As  tne  l°ad  moves  from  L0  to  L6,  RQ 
increases  uniformly  from  o  to  i  lb.;  and  MI  =  (I  —  a)R& 
increases  uniformly  from  o  to  /  —  a  ft.-lb.  Likewise  the  ordi- 
nate  to  the  line  QH ,  increasing  uniformly  from  o  at  L0  to  I  —  a 


FIG.  107. 

ft.-lb.  at  ZG,  represents  (/  —  O)RQ.  Hence,  the  line  QN  is  the 
influence  line  for  Mz,  when  the  load  of  i  lb.  is  between  L0 
and  L%. 

Let  PI  represent  the  load  on  the  segment  a  and  let  P  represent 
the  total  load  on  the  span;  then  the  criterion  for  a  maximum 
Mzt  developed  as  in  Article  87,  is 


The  stress  in  U\Uz  or  in  L2L3  for  either  truss,  found  by 
dividing  M^  by  the  proper  arm,  is  a  maximum  when  Mz  is  a 
maximum. 


SEC.  Ill 


BRIDGES 


This  criterion  is  often  expressed  in  the  form 

'  1    *-!>• 


or 


Pi 

a 


I-  a 


and  may  be  stated  as  follows:  A  maximum  stress  in  any  chord 
member  of  a  Pratt  or  Parker  truss;  or  of  any  other  truss  in 
which  the  influence  line  is  a  triangle  (as  in  Fig.  107),  occurs 
whenever  the  train  is  so  placed  that  the  average  load  per  foot  on 
each  of  the  two  segments  into  which  the  center  of  moments 
divides  the  span,  equals  the  average  load  per  foot  on  the  span. 
If  the  trusses  in  Fig.  107  have  six  equal  panels,  one-sixth  of 
the  total  load  on  the  span  is  placed  on  the  segment  o-i  for  maxi- 
mum stress  in  LJL*\  one- third  is 
placed  on  the  segment  0-2  for  maxi- 
mum stresses  in  UiU2  and  L2L3;  one- 
half  is  placed  on  segment  0-3  for 
maximum  stress  in  U^U^. 

100.  The  stress  in  a  web  member 
of  a  parallel  chord  truss  is  usually 
determined    by    passing    a    section 
through  the  web  member  and   two 
chord  members,  in  such   a   manner 
that  the  truss  is  cut  into  two  por- 
tions.    The  shear,  or  algebraic  sum,  of  the  vertical  forces  acting 
on  one  side  (either  side)  of  the  section,  equals  the  vertical  com- 
ponent of  the  stress  in  the  web  member. 

101.  Influence  Line  for  Shear  in  a  Panel. — The  criterion  for 
the  vertical  component  of  the  maximum  stress  in  U\L<i  (Fig. 
108)  will  be  developed  by  drawing  an  influence  line,  showing  the 
variation  of  the  shear  in  panel  1-2  as  a  load  of  i  Ib.  moves  across 
the  span. 

When  the  load  of  i  Ib.  is  between  Z,4  and  L%,  the  shear  in  the 
panel  equals  the  left  reaction;  which  increases  from  o  to  i  Ib.  as 
the  load  moves  from  L±  to  LQ.  Hence,  the  line  OT  is  the 
influence  line  for  shear  in  the  panel,  when  the  load  of  i  Ib.  is 
between  L4  and  L2;  and  TV  =  +%.  When  the  load  of  i  Ib. 


FIG.   108. 


152  THEORY  OF  FRAMED  STRUCTURES       CHAP.  IV 

is  between  L0  and  LI,  the  shear  in  the  panel  is  the  right  reaction 
taken  negatively.  Therefore  QS  is  the  influence  line  for  shear 
in  the  panel,  when  the  load  of  i  Ib.  is  between  L0  and  L\\  and 

us  =  -y±. 

When  the  load  of  i  Ib.  is  in  the  panel,  or  between  LI  and  L2, 
there  are  two  forces  acting  on  either  side  of  the  section.  On  the 
left,  there  are  the  truss  reaction  at  L0  and  the  floor  beam  load 
at  LI;  while  on  the  right  there  are  the  truss  reaction  at  L4  and 
the  floor-beam  load  at  L2.  Let  the  load  of  i  Ib.  be  in  the  panel 
and  at  a  distance  x  from  L2.  Consider  the  forces  on  the  left 
of  the  section,  and  let  R  represent  the  truss  reaction  at  L0,  r  the 
floor-beam  load  at  LI  and  F  the  shear  in  the  panel  1-2;  then 

D       50  +  x       ,  x 

R  =  —       -  and  r  =  - 
100  25 

F  =  R  -  r  =  5°_:^J* 
100 

This  equation,  being  of  the  first  degree,  may  be  represented  by  a 
straight  line.  When  the  load  of  i  Ib.  is  at  L2,  x  =  o,  and 
F  =  y%.  When  the  load  of  i  Ib.  is  at  LI,  x  =  25,  and  F  =  —  y±. 
Therefore  the  line  TS,  having  the  ordinates  TV  =  %  at  L2,  and 
US  =  —  Y±  at  LI,  is  the  influence  line  for  shear  in  the  panel; 
when  the  load  of  i  Ib.  is  between  L\  and  L2.  Hence,  the  line 
OTNSQ  is  the  influence  line  for  shear  in  the  panel  1-2. 

Positive  shear  in  panel  1-2  causes  tension  in  £/iL2.  As  the 
load  of  i  Ib.  moves  to  the  left  from  L4,  the  tension  in  U\Li 
increases  to  a  maximum  as  the  load  arrives  at  L2.  As  the  load 
moves  forward,  the  tension  decreases  rapidly,  and  the  stress  in 
£/iL2  becomes  zero  when  the  load  is  directly  over  N.  As  the 
load  passes  N,  the  member  resists  a  compressive  stress,  increas- 
ing to  a  maximum  as  the  load  reaches  Lr,  after  which  the  com- 
pression decreases  until  the  stress  becomes  zero,  as  the  load 
of  i  Ib.  leaves  the  span. 

102.  Criterion  for  Maximum  Shear  in  a  Panel. — The  shear 
in  panel  1-2  (Fig.  109)  may  be  determined  by  multiplying  each 
load  by  its  corresponding  ordinate  in  the  influence  line  diagram, 
and  finding  the  sum  of  these  products.  The  influence  line  is 
composed  of  three  lines  QS,  ST  and  TO\  and  the  loads  over  each 


SEC.  Ill 


BRIDGES 


153 


line  may  be  combined  respectively  into  PI,  and  P2  and  P3  as 
shown.  Let  y^  y*  and  ys  represent  respectively  the  ordinates 
under  the  centers  of  gravity  of  PI,  P2  and  P3;  and  let  F  repre- 
sent the  shear  in  the  panel;  then 

F  =  P3^3  ±  P23>2  -  Piyi 

The  ordinate  yz  will  have  a  plus  or  minus  sign  as  the  resultant 
P2  is  on  the  right  or  left  of  N. 

Now  let  the  loads  move  a  small  distance  d  to  the  left,  in  such 
a  way  that  no  loads  cross  the  panel  points  L0,  LI,  L2  or  L±. 


FIG.  109. 

Or  in    other  words,  PI,  P2  and  P3  are  to  remain  constant. 

The  change  in  shear  in  the  panel  is 

AF  =  P3;y33  -  P2}>22  +  PI^U. 

If  yz  had  been  a  negative  ordinate,  it  would  have  been 
lengthened  by  the  movement,  an  amount  ;y22.  In  either  case 
the  shear  is  decreased  and  ;y22  is  negative.  Is  this  change  in 
F,  caused  by  a  movement  of  the  loads  to  the  left,  a  positive  or  a 
negative  quantity?  The  answer  depends  upon  whether 


for  as  long  as  P2;y22  <P3;y33  -f 

the  change  in  F,  caused  by  a  movement  of  the  loads  to  the  left, 

will  be  a  positive  quantity. 


154  THEORY  OF  FRAMED  STRUCTURES       CHAP.  IV 

From  similar  triangles 

yu:d::l:ioo 
y22:d::%:2$ 
yzs:d::l:  100 

d  ^d  d 

or  yu  =       —  ,  3/22  =  --  and  y3S  =  --- 

100  100  100 

Hence  the  change  in  F,  caused  by  a  movement  of  the  loads  to 
the  left,  will  be  a  positive  quantity  as  long  as 


IOO  100  100 

or  as  long  as  ^P2  <Ps  +  PI. 

Likewise  the  change  in  F,  caused  by  a  movement  of  the  loads 

to  the  right  will  be  a  negative  quantity  so  long  as 

3P2  <  P3  +  Pi. 

Let  Pi  -f  P2  +  P3  =  P  =  total  load  on  the  span; 
if  3P2<P3  +  Pi 

and  P2  =  P2 

then  4P2  <P 

or  P2  <P 

4 

Hence,  a  positive  shear  in  panel  1-2  will  be  increased  by  a  move- 
ment of  the  loads  to  the  left,  and  a  negative  shear  will  be  increased 
by  a  movement  of  the  loads  to  the  right  so  long  as 


Hence,  the  criterion  for  maximum  positive  or  negative  shear 
in  panel  1-2  is 

AS<       .  .-  ; 

For  a  maximum  positive  shear  the  influence  line  indicates 
that  as  many  wheels  as  possible  should  be  on  the  span  between 
L4  and  L2,  and  a  few  wheels  (not  exceeding  one-fourth  of  the 
total  load)  should  move  across  L2  into  the  panel.  For  a  maxi- 
mum negative  shear,  as  many  wheels  as  possible  should  be  on 
the  span  between  LQ  and  LI,  and  a  few  loads  (not  exceeding 


SEC.  Ill  BRIDGES  155 

one-fourth  of  the  total  load)  should  move  across  LI  into  the 
panel. 

Suppose  that  the  panel  lengths  in  Fig.  109  are  changed  so 
that  LoLi  =  20  ft.,  LiL2  =  25  ft.,  L2L3  =  38  ft.  and 
17  ft.     Will  this  change  modify  the  criterion 


and  if  so,  how? 
103.  Illustrative  Problems. 

What  is  the  maximum  tension  in  £/iL2  (Fig.  109)  for  an  £-40 
loading?     The  maximum  compression? 

Tension.  —  Move  the  train  on  the  span  from  the  right  end 
until  wheel  2  is  at  L2.  Wheel  10  is  2  ft.  on  the  span  and  the 
total  load  is  152. 

4)152 


When  wheel  2  is  approaching,  or  has  passed  Z2,  the  load  in  the 
panel  is  less  than  one-fourth  the  total  load  on  the  span;  con- 
sequently the  shear  in  the  panel  is  increasing  in  either  case. 
Try  wheel  3  at  L2. 

4)-^      ,0 

38^0  O.K- 

Wheel  3  satisfies  the  criterion  for  maximum  positive  shear  in 
the  panel,  for  the  load  in  the  panel  is  less  than  one-fourth  the 
total  load  when  wheel  3  is  approaching  L2,  and  the  shear  is 
increasing;  also  the  load  in  the  panel  is  greater  than  one-fourth 
the  total  load  when  wheel  3  has  passed  L2,  and  the  shear  is 
decreasing.  Make  a  sketch  of  the  truss  showing  the  train  with 
wheel  3  at  L2. 

The  shear  in  the  panel  when  wheel  3  is  at  L2  is  determined  by 
finding  the  algebraic  sum  of  the  vertical  forces  acting  on  one 
side  (either  side)  of  a  section  through  the  panel.  There  are 
four  forces  acting  on  the  right,  i.e.,  the  truss  reaction  at  L4 
and  the  three  floor  beam  loads  at  L2,  L3  and  £4  respectively. 
There  are  only  two  forces  acting  on  the  left  of  the  section, 
i.e.,  the  truss  reaction  at  LQ  and  the  floor  beam  load  at  L\. 
The  shear  in  the  panel  will  therefore  be  computed  by  considering 


156  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

the  forces  acting  on  the  left  of  the  section.     Wheel  10  is  7 
ft.  on  the  span. 

4,632 

152  X  7  =  1,064 
100)5,696 

56.96  =  left  reaction 

2  "?O 

-^-  =    9.2     =  floor-beam  load  at  LI 

2  £ 

47.76  =  shear  in  the  panel. 
The  length  of  the  member  £7iL2  is  40.6  ft. 

40.6 

47  -7^  X  -  =  60.6  =  the  maximum  tension  in  £/iL2. 
o 

Compression.  —  Move  the  train  on  the  span  from  the  left  until 
wheel  2  is  at  Li.  Wheel  6  is  i  ft.  on  the  span  and  the  total 
load  is  103. 

4)103 


Wheel  2  satisfies  the  criterion  for  maximum  negative  shear  in 
the  panel,  for  the  load  in  the  panel  is  less  than  one-fourth  the 
total  load  when  wheel  2  is  approaching  Li,  and  the  negative 
shear  is  increasing;  also  the  load  in  the  panel  is  greater  than  one- 
fourth  the  total  load  when  wheel  2  has  passed  LI,  and  the  nega- 
tive shear  is  decreasing.  Make  a  sketch  of  the  truss  showing 
the  train  with  wheel  2  at  L\. 

There  are  two  forces  acting  on  the  right  of  the  section  i.e., 
the  truss  reaction  at  L±  and  the  floor  beam  load  at  L^.  Wheel 
6  is  i  ft.  on  the  span. 

1,640 

103  X  i  =      103 
100)1,743 

17.43    =  right  reaction 
—  =       3.2       =  floor  beam  load  at  Lz 

14.23     =  negative  shear  in  the  panel 

14.23  X  -  =  18.1  =  the  maximum  compression  in  U\Li. 
The  member  UiL^  may  have  a  maximum  tension  of  60.6, 


SEC.  Ill 


BRIDGES 


or  a  maximum  compression  of  18.2.  It  is  clear  that,  if  the 
member  U\L<i  were  removed  and  a  member  L\V^  substituted, 
the  member  L\Ui  would  have  a  maximum  tension  of  18.2 
and  a  maximum  compression  of  60.6. 

UiLi. — This  member  supports  the  floor  beam  load  at  Li, 
consequently  the  stress  is  a  maximum  when  the  floor  beam  load 
is  a  maximum.  As  a  load  of  i  Ib.  moves  from  L±  to  Z,2,  the 
floor  beam  load  is  zero.  The  floor  beam  load  increases  uni- 
formly from  o  to  i  Ib.  as  the  load  moves  from  Li  to  LI,  and 
decreases  uniformly  from  i  Ib.  to  o  as  the  load  moves  from  LI 
to  LQ.  The  influence  line  diagram  for  the  stress  in  U\L\  is 


FIG.  no. 

an  isosceles  triangle  having  a  base  50  ft.  long,  and  an  altitude 
of  i  Ib.  under  L\.  The  criterion  for  maximum  stress  in  U\L\, 
developed  from  this  influence  line,  shows  that  the  panels 
LoL2  should  be  loaded  with  one-half  the  load  in  each  panel. 
Either  wheel  4  or  wheel  13  at  LI  satisfies  this  criterion;  the  stress 
being  the  same  in  either  case.  When  wheel  4  is  at  LI,  wheel  8 
is  at  L2  and  wheel  i  is  7  ft.  from  L0.  The  stress  in  UiLi  which 
equals  the  floor  beam  load  at  LI  is  75.6  tension. 

104.  General  Criterion  for  Maximum  Shear  in  a  Panel. — 
Let  F  represent  the  shear  in  panel  2-3  of  the  parallel  chord 
truss  in  Fig.  no,  when  the  truss  supports  any  system  of 
vertical  loads.  Since  F  represents  also  the  vertical  component 
of  the  stress  in  C/aLs,  it  is  clear  that  the  stress  is  a  maximum 


158  THEORY  OF  FRAMED  STRUCTURES       CHAP.  IV 

when  the  shear  in  the  panel  is  a  maximum.  The  influence  for 
shear  in  panel  2-3  has  been  drawn  as  in  Article  101.  If  PI 
represents  the  load  on  the  segment  a;  P^  the  load  in  the 
panel  1-2 ;  PS,  the  load  on  the  segment  c ;  and  P  the  total  load 
on  the  span,  the  criterion  for  maximum  shear,  which  may 
be  developed  in  the  same  manner  as  in  Article  102,  is 

P   <  b  P 
2  >  I 

This  criterion  is  not  a  function  of  either  a  or  c,  and  conse- 
quently is  applicable  to  any  panel  of  the  truss;  when  P2  is  the 
load  in  the  panel,  P  is  the  total  load  on  the  span,  b  is  the  length 
of  the  panel  and  /  is  the  length  of  the  span.  The  criterion  is 
often  expressed  in  the  form 

•  ,    ,._», 

PI    P 

T  =  T 

and  may  be  stated  as  follows:  The  maximum  stress  in  any 
diagonal  of  a  parallel  chord  truss  (Pratt  or  Warren)  occurs 
when  the  load  in  the  panel  divided  by  the  panel  length,  equals 
the  load  on  the  span  divided  by  the  span  length. 

If  the  truss  in  Fig.  no  contains  eight  equal  panels,  it  is  clear 
that  one-eighth  of  the  total  load  on  the  span  should  be  placed  in 
any  panel  for  maximum  stress  in  the  diagonal  of  that  panel; 
the  train  approaches  from  the  right  for  positive  shear  and  from 
the  left  for  negative  shear. 

In  a  through  bridge,  having  the  floor  beams  at  the  bottom 
chord,  the  maximum  stress  in  a  vertical  member,  UzL^  for 
example,  equals  the  maximum  shear  in  panel  2-3,  for  that 
portion  of  U^Lz  between  the  top  of  the  floor  beam  and  U%.  The 
stress  in  that  part  of  UzLi  (if  any  exists)  between  the  bottom 
of  the  floor  beam  and  the  end  connection  differs  from  the  stress 
above  the  floor  beam  by  the  amount  of  the  floor  beam  load. 

From  similar  triangles 

c      a  -\-  c 


I  '       I 


NV':b 


SEC.  Ill  BRIDGES  159 

therefore  NV  :  c  :  :  b  :  a  +  c 

NV:VO  ::b:a  +  c 

NViNV  +  VO  ::b  :b  +  a  +  c 


_ 
NO  ~"  1 

P*       P 

but  since  ~r    =  T 

o          I 

P*        P 
NV^NO 

Thus  it  is  clear  that,  when  the  right  portion  of  the  span  is 
loaded  for  maximum  positive  shear  in  any  panel  ;  the  load  in  the 
panel  divided  by  the  length  NV  equals  the  total  load  in  the 
span  divided  by  the  length  NO.  In  other  words,  if  the  truss 
has  n  equal  panels  and  the  influence  line  OTNSQ  is  drawn  for 
shear  in  any  panel 

NV       i 
NO~  n 

Similarly  it  may  be  shown  that 

NU       i 
NQ  ~  n 

The  influence  line  NTO,  for  positive  shear  in  any  panel,  has 
the  same  general  characteristics  as  an  influence  line  for  bending 
moment  at  the  point  F  in  a  beam  having  the  length  NO;  the 
criterions  are  the  same  and  the  positions  of  the  train  are  the 
same  in  both  cases.  The  position  of  the  train  for  maximum 
negative  shear,  in  any  panel,  is  the  same  as  for  maximum  bend- 
ing moment  at  U  in  a  beam  having  the  length  NQ. 

105.  Impact.  —  In  computing  live  load  stresses  in  a  bridge, 
the  train  is  considered  as  a  static  load,  gently  placed  upon  the 
structure  in  the  required  position  for  maximum  stress  in  a 
given  member.  In  order  to  provide  for  the  increased  stress 
caused  by  dynamic  effect  of  the  train  in  motion,  an  additional 
stress  known  as  impact  is  combined  with  each  live  load  stress. 
The  amount  of  impact  to  be  added  is  determined  arbitrarily 
from  an  empirical  formula.  The  one  most  frequently  used  is 

T  -  T        3°° 

L     —    ,L/  ,      , 

300   +   / 


i6o 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  IV 


in  which 

7  =  impact  stress 

L  =  live  load  stress 

/  =  length  of  train  causing  the  live  load  stress. 
For  example,  the  live  load  stress  in  U\Uz  (Fig.  106),  as  deter- 
mined in  Article  98,  is  100.6.     The  impact  stress  is 

100.6  X  -: =  75.4  compressive 

300  +  100 

The  live  load  tension  stress  in  U\L^  (Fig.  108),  as  determined 

Ui         U2         U3         U4         Us         U&        U7 


—L 
I* 


t     L'     Lj     L>     I  ,  Ls     I     Ll 

[^.. QQ)25^?00 >J 

FIG.  in. 

in  Article  103  is  60.6.     The  impact  stress  is 

60.6  X — . =  50.2  tensile 

300  +  63 

The  live  load  compression  stress  is  18.1,  and  the  impact  is 

18.1  X  — : =  16.3  compressive 

300  +  33 

The  live  load  stress  in  U\L\  is  75.6.  The  length  of  the  train 
causing  this  stress  extends  over  two  panels  only,  hence  the 
impact  stress  is 


75.6  X 


=  66.1 


300  -1-  43 

106.  Stresses  in  a  2oo-ft.  Pratt  Truss. — (Fig.  in).  The 
assumed  dead  loads  are  as  follows: 

Track  (rails  and  ties) 450 

Floor  (floor  beams  and  stringers) 550 

Two  trusses  and  bracing i ,  200 

2,200  Ib.  per  linear  foot 

The  live  load  is  a  Cooper's  £-40  train.  The  impact  will  be 
taken  in  accordance  with  the  formula  of  Article  105. 

Dead-load  Stresses. — The  dead  load  supported  by  each  truss 
is  1,100  Ib.  per  linear  foot,  or  27.5  per  panel.  The  stresses 


SEC.  Ill  BRIDGES  l6l 

may  be  determined  by  placing  a  load  of  27.5  at  the  seven  joints 
LI  to  LI  and  drawing  a  stress  diagram;  but  in  the  case  of 
parallel  chords  the  algebraic  method  is  considerably  shorter. 
One-half  the  load  per  linear  foot  on  each  truss  is  550  Ib.  (see 
Article  59) . 
0.550  X  i  X  25  X  7  X  25  =  i  6g 

35 

9.82  X  2  X  6  =  117.8  = 

9.82  X  3  X  5  =  143-7  = 

9.82  X  4  X  4  =  157.1  =  U*Ut 

The  length  of  each  diagonal  is  43  ft.  The  shear  in  panel  3-4  is 
one-half  a  panel  load,  or  13.75 

13-75  X  ^  =    i6-.9   = 

16.9  X  3    =    50.7  = 

16.9  X  5    =    84-5  = 

16.9  X  7    =  118.3  = 

13.75  = 

13-75  X  3    =    41-3  = 

27-5  : 

o  = 

It  is  more  accurate  to  consider  one-half  the  weight  of  the 
truss  as  concentrated  at  the  top  chord  panel  points.  This 
would  affect  the  stresses  in  the  vertical  members  only,  increas- 
ing the  stress  in  UzLz,  U^Lz  and  UtL±  and  decreasing  the  stress 
in  UiLi  by  7.5 

Live  load  and  Impact  Stresses 
LoL2  Wheel  4  at  LI 

16,364 

284  X  84  =  23,856 
842  =     7,056 

8)47,276 


480. 


35)5,429-5 
LL  =  155.1 

/  =  155.1  X  ^  =  94-5 
ii 


1 62  THEORY   OF   FRAMED    STRUCTURES  CHAP.  IV 

UiUz  and  L2L3  Wheel  7  at  L2 
16,364 

284  X  78  =  22,152 

782  =    6,084 

4)44,600 

11,150  ••: 

2,155 


/=  257X^=158.2 

UzU3  and  L3L4  Wheel  n  at  Ls 

16,364 

284  X  80  =  22,720 
852  =  6,400 

45,484 
o.375 


17,056-5 
5,848 

35)11,208.5 
LL  =  320.2 

1  =  320'2  X          =  '96-5 


Wheel  13  at  L4 

16,364 

284  X  65  =  18,460 
652  =    4,225 

2)39,049 

19,524-5 
7,668 


35)11,856.5 
LL=  338.8 

7  =  338.8  X      °  =  214.5 


SEC.  Ill  BRIDGES  163 

The  stresses  in  all  chord  members  of  the  left  half  of  the  truss 
are  a  maximum  for  positions  of  the  train  when  crossing  the  span 
from  right  to  left.  This  is  not  a  general  statement,  to  be 
accepted  for  all  trusses,  although  it  is  true  for  this  truss.  The 
chord  members  of  the  right  half  will  have  their  maximum 
stresses  when  the  train  crosses  from  left  to  right.  The  stresses 
for  all  web  members  will  be  computed  for  the  train  moving 
from  right  to  left.  The  length  of  any  diagonal  member  is  43  ft. 
Hence  if 

v  =  the  vertical  component  of  the  stress  in  a  diagonal 
and     s  =  the  stress  in  the  diagonal 

then    s  =  --  X  43 

O  0 

or       s  =  1.232; 


Wheel  2  at  L7 
1,640 

103  X  i  =      103 

200)1,743 

8.7 

—  - 

25       5^5  X  1.23  =  6.8  =  LL 


U&LQ  and  L5Ut  Wheel  2  at  L6 
4,632 

152   X    2    =        304 
200)4,936 
24.7 

3-2 


21.5=  LL 

-«:  -/ 


164  THEORY    OF   FRAMED    STRUCTURES  CHAP.   IV 

U5L5  and  L,U5  Wheel  2  at  L5 

8,728 
232  X  4  = 


200)9,656 

48-3 
3-2 
45.1  =  LL 

=  35-3   =/ 

45.1  X  1.23  =  55.4  =  LL 
35.3  X  1.23  =  43.4  =  / 


and  *73L4  Wheel  3  at  L4 

16,364 

284  X  4  =     i,i36 

42  =  _  16 

200)17,516 

87.6 


78.4  =LL 

200 

78.4X^  =  57.  =/ 

78.4  X  1.23  =  96.4  =  LL  ]  \      ^ 
57.     X  1.23  =  70.1  =  I     I 


i  and  £7^3  Wheel  3  at  L3 

16,364 
284  X  29  =  8,236 

292  =        841 
200)25,441 


tf,t, 

L\ 
80.9  X  1.23  =    99.6  =  I 


118.     X  1.23  =  145.2  =  LL  ^  v  L 


SEC.  Ill  BRIDGES  165 

UiL2  Wheel  3  at  L2 

16,364 

284  X  54  =  15,336 
542  =     2,916 

200)34,616 


9.2 


163.9  X  1.23  =  201.6  =  LL 


201.6  X  =  130.6  =  / 

463 


L»Ui  Wheel  4  at  Li 
16,364 

284  X  84  =  23,856 

842=     7.056 

200)47,276 

236.4 
480 


217.2  X  1.23  =  267.2  =  LL 

267.2  X  —  «=  162.5  =  I 
493 

tfiLi  Wheel  4  at  Ll 

Moment  of  wheels  i  to  8  about  8  =  2851 
Moment  of  wheels  i  to  4  about  4  =     480 
2,851 

480  X    2    =     960 
25)1,891 

75.6  =  LL 

75.6  X  —  =  66.1  =  7 
343 

When  the  train  crosses  the  span  in  one  direction,  it  is  obvious 
that  the  stress  in  any  web  member  of  the  left  half  is  the  same  as 
the  stress  in  the  corresponding  member  of  the  right  half,  when 
the  train  crosses  in  the  opposite  direction.  Considering  one- 
half  of  the  structure,  we  shall  have  the  various  combinations  of 


i66 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


stresses,  as  listed  in  Fig.  112.  When  the  dead  and  live  load 
stresses  are  opposite  in  character,  only  two-thirds  of  the  dead- 
load  stress  is  considered.  Thus  in  U^L^  the  total  stress  is 
+  183.4  when  the  train  crosses  from  right  to  left,  and  —87.5 
when  the  train  crosses  from  left  to  right. 

In  UiLz  the  dead-load  tensile  stress  is  greater  than  the  live- 


u' 


D=-II7.S 

L  -  -2510 

I ---158.2 

-533.0 


FIG.  112. 


U,  -533  U?          -560.4          Ik          -7/0.4 


L,          *3/8.3  lz          +533  L3          ^660.4  L4 

FIG.   113. 


load  and  impact  compressive  stresses,  and  there  is  no  reversal. 
107.  Counters. — The  diagonals  in  two  of  the  panels  in  Fig. 
112  must  be  designed  to  resist  both  tensile  and  compressive 
stresses.  If  eye-bars  or  other  flexible  members,  which  are  not 
capable  of  taking  compressive  stress,  are  used;  it  will  be  neces- 
sary to  introduce  counters  in  panels  2-3  and  3-4,  as  shown  by 
the  dotted  lines  in  Fig.  113. 


SEC.  Ill  BRIDGES  167 

A  counter  carries  no  stress,  except  when  the  shear  in  the  panel 
is  of  such  a  nature  that  the  stress  in  the  main  diagonal  would  be 
compressive  if  the  counter  were  omitted.  The  tensile  stress 
carried  by  the  counter  in  a  parallel  chord  truss,  obviously  has 
the  same  magnitude  as  the  compressive  stress  in  the  main 
diagonal  if  no  counter  were  used.  Thus  in  Fig.  113  the  maxi- 
mum stress  in  UsLi  is  +  183  .  4,  and  the  maximum  stress  in  LzUi 
is  +87.5.  The  maximum  compressive  stress  in  U^L*  has  the 
same  magnitude  as  the  maximum  tensile  stress  in  UzLz,  when 
no  counters  are  used.  When  the  bridge  is  loaded  for  maximum 
chord  stresses,  the  counters  are  not  in  action  and  consequently 
have  no  effect  upon  the  maximum  chord  stresses. 

The  Pratt  truss  in  Fig.  112  may  be  transformed  into  a  War- 
ren truss,  by  substituting  the  diagonal  LzUz  for  the  diagonal 
U%Lz.  The  stress  in  L^Us  will  be  —  295  .  5  or  +  14  .  8  ;  the  stress 
in  U2Uz  will  be  the  same  as  in  UiU^  the  stress  in  LzLz  will  be  the 
same  as  in  L$L±,  the  stress  in  U^L^  will  be  the  same  as  in  U±L±\ 
and  the  stress  in  UzLz  will  be  the  same  as  in  UiL\. 

108.  Truss  with  an  Odd  Number  of  Panels.  —  The  bottom 
chord  in  the  center  panel  of  a  truss  having  an  odd  number  of 
panels  requires  a  special  consideration.  The  truss  in  Fig.  114 
is  loaded  for  the  maximum  moment  at  L4.  The  moment  at 
L4  is  9,088.57  ft.-lb.,  and  the  moment  at  £3  is  8,929.43  ft.-lb. 
Since  the  moment  at  L4  is  greater  than  the  moment  at  L3,  the 
shear  in  the  panel  3-4  is  positive,  and  the  diagonal  U^Li  is 
acting.  The  shear  in  panel  3-4  equals  the  left  reaction,  minus 
the  loads  from  wheel  i  to  wheel  9  inclusive,  minus  that  part 
of  the  load  in  the  panel  which  is  carried  at  L3,  or 

F  =  167.57  -  J42  -  19-2  =  6.37 


The  shear  may  also  be  found  by  dividing  the  difference  in 
moments  at  L4  and  Lz  by  the  panel  length,  thus 

p  =  9,088.57  -  8,929.43  = 

The  truss  in  Fig.  115  is  loaded  for  the  maximum  moment 
at  LZ.  The  moment  at  LZ  is  9,155.86  ft.-lb.,  and  the  moment 
at  L4  is  8,944.14  ft.-lb.  Since  the  moment  at  L3  is  greater  than 


i68 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


the  moment  at  Z,4,  the  shear  in  the  panel  3-4  is  negative  and 
the  diagonal  Z/3Z74  is  acting.     The  shear  in  panel  3-4  is 

F  =  200.05  —  172  —  36.52  =  —8.47 
p  =  ^,944.14  -  9,155.86  =  _g 


or 


PIG.  115. 


PIG.  116. 


The  maximum  stress  in  UzUi  is  9,155.86/34  =  —269.3. 
This  stress  occurs  when  the  train  crosses  the  span  from  right 
to  left  with  wheel  1 1  at  L$ ;  or  when  the  train  crosses  from  left 
to  right  and  wheel  1 1  is  at  L4.  Hence  the  maximum  stress  in 
UzUz,  UzU*  and  U*U6  is  -269.3. 


SEC.  IV  BRIDGES  169 

The  stress  in  L*L±  is  obtained  by  dividing  the  moment  at 
Uz  in  Fig.  114,  or  the  moment  at  Z74  in  Fig.  115,  by  34;  but  in 
either  case  the  moment  at  the  other  end  of  the  panel  is  greater. 
It  is  clear,  therefore,  that  the  maximum  stress  in  L3L4  occurs 
when  the  moments  at  Ls  and  L4  are  equal,  or  when  the  shear 
in  the  panel  is  zero  and  neither  diagonal  is  acting. 

The  total  load  on  the  span  in  Fig.  114  is  364. 

364  _. 

T    " 

The  load  in  panel  3-4  is  less  or  greater  than  one-seventh  of  the 
load  on  the  span  as  wheel  13  is  crossing  Z,4,  therefore  the  shear 
in  the  panel  is  a  maximum  (not  the  maximum  to  be  sure,  but 
still  a  maximum).  There  are  three  intermediate  critical 
positions  of  the  train  between  those  shown  in  Figs.  114  and 
115 — wheel  14  at  L4,  wheel  10  at  Z,3  and  wheel  15  at  Z,4 — and  for 
each  position  the  shear  in  panel  3-4  is  decreasing,  for  the  load 
in  the  panel  in  each  case  is  greater  than  one-seventh  of  the 
load  on  the  span.  Hence,  as  the  train  moves  from  the  position 
in  Fig.  114  to  the  position  in  Fig.  115,  and  the  shear  in  panel 
3-4  decreases  from  +6.37  to  —8.47;  there  is  one,  and  only  one, 
position  of  the  train  for  which  the  shear  in  the  panel  is  zero. 

In  the  present  case  the  shear  in  panel  3-4  is  (approximately) 
zero  when  wheel  10  is  at  L$  (Fig.  116)  and  the  moment  at 
either  U^  or  £/4  is  9,048.5  ft.-lb.  Hence,  the  maximum  stress 
in  LzL±  is  +266.1.  This  stress  is  slightly  less  than  the  stress 
in  UzUtj  and  in  practice  is  usually  assumed  the  same  as  the 
stress  in  U^Ui. 

SEC.  IV.     PARKER  TRUSSES 

109.  Stress  in  Web  Member  of  Parker  Truss. — It  was 

shown  in  Article  99  that  the  criterion  for  maximum  stress  in  any 
chord  member  of  the  Parker  truss  (Fig.  107)  is  the  same  as  for 
the  corresponding  member  of  the  Pratt  truss.  It  is  true  also 
that  the  criterion  for  maximum  shear  in  any  panel  of  the 
Parker  truss  is  the  same  as  for  the  corresponding  panel  in  the 
Pratt  truss.  In  a  consideration  of  the  maximum  stresses  for 
web  members,  a  clear  distinction  should  be  made.  In  the 


1 7o 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.   IV 


Pratt  truss,  the  stress  in  any  diagonal  is  proportional  to  the 
shear  in  the  panel;  and  therefore  is  a  maximum  when  the  shear 
in  the  panel  is  a  maximum.  In  the  Parker  truss,  the  stress 
in  any  diagonal  C/iL2,  for  example,  is  not  proportional  to  the 
shear  in  the  panel,  for  the  vertical  component  of  the  stress  in 
UiUz  enters  into  the  solution. 
no.  Influence  Line  for  Web  Member  of  Parker  Truss.— If 


FIG.  117. 


the  truss  in  Fig.  117  is  loaded  in  any  manner,  the  stress  in 
may  be  determined  by  passing  a  section  through  the  panel  2-3, 
and  balancing  the  moments  of  all  the  forces  acting  on  one  side 
(either  side)  of  the  section  about  the  intersection  of  the  two 
chord  members  cut  by  the  section.  Let  MI  represent  the  alge- 
braic sum  of  the  moments  about  /  of  all  the  forces  acting  on 
one  side  (either  side)  of  the  section  through  the  panel  2-3,  as  a 
load  of  i  Ib.  moves  across  the  span. 

When  the  load  of  i  Ib.  is  between  L8  and  L$,  the  only  force 


SEC.  IV  BRIDGES  171 

acting  on  the  left  of  the  section  is  the  left  reaction  R0,  and  MI  = 
iSoRQ.  As  the  load  moves  from  Z,8  to  L0,  R0  increases  uni- 
formly from  o  to  i  lb.,  and  iScxfto  increases  uniformly  from  o 
to  1 80  ft.-lb.  Likewise  the  ordinate  to  the  line  OK,  increasing 
uniformly  from  zero  at  O  to  180  ft.-lb.  at  Q,  represents  iSoR0. 
Hence,  the  line  OT  is  the  influence  line  for  Mz  —  i8o7?o,  when 
the  load  of  i  lb.  is  between  L8  and  L3.  Ml  for  the  force  acting 
on  the  left  of  the  section  is  counter-clockwise  about  /,  and  for 
this  reason  QK  is  laid  off  above  the  line  OQ. 

When  the  load  of  i  lb.  is  between  Z,0  and  L^  the  only  force 
acting  on  the  right  of  the  section  is  the  right  reaction  R$  and 
Mj  =  42oR8.  As  the  load  moves  from  LQ  to  L8,  R&  increases 
uniformly  from  o  to  i  lb.  and  420^3  increases  uniformly  from 

0  to  420  ft.-lb.     Likewise  the  ordinate  to  the  line  QH,  increasing 
uniformly  from  zero  at  Q  to  420  ft.-lb.  at  0,  represents  42cxR8. 
Hence  the  line  QS  is  the  influence  line  for  M\,  when  the  load  of 

1  lb.  is  between  L0  and  L2.     MI  for  the  force  acting  on  the  right 
of  the  section  is  counter-clockwise,  consequently  Mx  for  the 
forces  acting  on  the  left  of  the  section  is  clockwise,  and  for  this 
reason  OH  is  laid  off  below  the  line  OQ. 

TV  =  1 80  X  |  =  112.5 
8 

2 

US  =  —420  X  -  =  —105 
8 

When  the  load  of  i  lb.  is  between  LI  and  L3,  there  are  two 
forces  acting  on  either  side  of  the  section.  On  the  left,  there 
are  the  truss  reaction  R0  at  L0,  and  the  floor-beam  load  r^  at 
L2;  on  the  right  of  the  section,  there  are  the  truss  reaction  R$ 
at  L8,  and  the  floor-beam  load  r%  at  L3.  Let  the  load  of  i  lb.  be 
in  the  panel  2-3  at  a  distance  x  from  Z,3,  and  consider  the  forces 
acting  on  the  left  of  the  section. 

*0  „  '50  +  *  rz  =  _* 
240  30 

Mi  =  iScxRo  —  2407-2 

,,  180(150  +  x)  240X 

Ml  —  -  —  -— — - 

240  30 


172  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

This  equation,  being  of  the  first  degree,  may  be  represented 
by  a  straight  line.  When  the  load  of  i  Ib.  is  at  £3,  x  =  o  and 
M i  =  112.5.  When  the  load  of  i  Ib.  is  at  L2,  x  =  30,  and  MT 
—  —105.  Therefore,  TS,  having  the  ordinates  TV  =  112.5 
at  Z,3,  and  US  =  — 105  at  L2,  is  the  influence  line  for  Mi  when 
the  load  of  i  Ib.  is  in  the  panel  2-3.  The  stress  in  U^Ls  is 
Mj  -f-  216.  The  stress  caused  by  the  load  of  i  Ib.  in  any  posi- 
tion, may  be  found  by  dividing  the  corresponding  ordinate  in 
the  influence  line  OTNSQ  by  216.  As  the  load  of  i  Ib.  moves 
to  the  left  from  L8,  the  tension  in  Z72Z,3  increases  to  a  maximum, 
as  the  load  arrives  at  L3.  As  the  load  moves  forward,  the  ten- 
sion decreases  rapidly  and  the  stress  in  UzLs  becomes  zero, 
when  the  load  arrives  at  the  point  directly  above  ^V.  As  the 
load  passes  N,  the  member  resists  a  compressive  stress  increas- 
ing to  a  maximum  as  the  load  arrives  at  L2;  after  which  the 
compression  decreases  to  zero,  as  the  load  leaves  the  span  at 

u 

Since  IL0:ILS::KQ:OH 

it  is  evident  that  the  two  lines  OK  and  HQ  produced,  intersect 
at  /  on  the  vertical  through  7.  This  influence  resembles  some- 
what the  influence  line  for  shear  in  the  panel  2-3.  If  the  length 
of  either  member  Z72L2  or  U^L^  were  changed  so  that  UiU% 
approached  a  direction  parallel  to  LzLsy  the  distance  between 
/  and  LQ  would  approach  infinity,  and  the  lines  KO  and  QH 
would  approach  a  condition  of  being  parallel. 

in.  Criterion  for  Maximum  Stress  in  Web  Member  of 
Parker  Truss. — The  member  U2L3  (Fig.  117)  will  be  used  as 
an  example.  A  train  is  assumed  to  be  on  the  span.  The 
stress  in  £72L3  is  M1  -f-  216.  The  value  of  M£  may  be  deter- 
mined by  taking  the  sum  of  the  products  of  each  load  and  its 
corresponding  ordinate  in  the  influence  line.  The  loads  will 
be  combined  into  three  groups  PI,  P2  and  P3,  corresponding  to 
the  three  portions  of  the  influence  line  QS,  ST  and  TO.  If 
yi,  yz  and  ys  represent  respectively  the  ordinates  under  the  cen- 
ters of  gravity  of  PI,  P2  and  P3,  then 

Mj  =  Pzyz  ±  P2;y2  -  P&I. 
Let  the  loads  move  a  small  distance  d  to  the  left  in  such  a 


SEC.  IV  BRIDGES  173 

way  that  no  loads  cross  the  panel  points  L0,  L2,  L3  and  L8.  Or 
in  other  words  PI,  PI  and  P%  are  to  remain  constant.  Is  this 
change  in  Mt,  caused  by  a  movement  of  the  loads  to  the  left, 
a  positive  or  a  negative  quantity?  The  answer  depends  upon 
whether 


for  as  long  as  ^2^22  <Pzy^  +  P\y\\ 

the  change  in  MT  caused  by  a  movement  of  the  loads  to  the  left, 
will  be  a  positive  quantity. 
From  similar  triangles 

yn'.d:  1420:240 

yn:d\  1217.  5:30 

yzz'd:  1180:240 

nd  2gd  id 

or  yn  =   J-  •  ?22  =  -    -   and  ;y33  =  - 

44  4 

Hence  the  change  in  Mz,  caused  by  a  movement  of  the  loads  to 
the  left,  will  be  a  positive  quantity  as  long  as 


4  44 

or  as  long  as  29P2  <^Ps  +  7  PI 

Likewise,  the  change  in  Mn  caused  by  a  movement  of  the  loads 
to  the  right,  will  be  a  negative  quantity  as  long  as 

29P2<3P3  +  7-Pi- 

The  tensile  stress  in  U^LS  is  a  maximum,  when  MI  has  a 
maximum  positive  value,  or  when  the  right  portion  of  the  span 
is  loaded.  As  the  loads  come  onto  the  span  at  L9  and  move  to 
the  left,  the  tension  in  UiLz  increases  as  long  as 


Let         Pi  +  P2  +  P3  =  P  the  total  load  on  the  span. 
If  29P2 

and          3P2 

then       3  2P2 

or  P2<    3_p+ip 

32          8 

Hence  the  criterion  for  maximum  tensile  stress  in  U^Lz  is 

P^-P  +  *A. 

32          8 


174  THEORY    OF    FRAMED    STRUCTURES  CHAP.   IV 

This  criterion  will  be  satisfied  when  the  critical  load  is  at  L3, 
and  a  few  loads  are  in  panel  2-3.  In  the  case  of  all  ordinary 
train  loads  the  criterion  will  be  satisfied  before  any  loads  pass 
L2;  in  which  case  PI  =  o  and  the  criterion  for  maximum  tensile 
stress  in  Z72L3  becomes 


32 

The  compressive  stress  in  Z72L3  is  a  maximum  when  M1  has 
a  maximum  negative  value,  or  when  the  left  portion  of  the  span 
is  loaded.  As  the  loads  come  onto  the  span  at  L0  and  move  to 
the  right  the  compression  in  £72L3  increases  as  long  as 


If  29P2<3P3 

7A  =  7^ 
then 


or  P2  <  -^  P  -  -  P3 

36  9 

Hence  the  criterion  for  maximum  compressive  stress  in  UiLz  is 


, 

36          9 

This  criterion  will  be  satisfied  when  the  critical  load  is  at  L2, 
and  a  few  loads  are  in  panel  2-3.  In  the  case  of  ordinary  train 
loads  the  criterion  will  be  satisfied  before  any  loads  pass  L3; 
in  which  case  P3  =  o  and  the  criterion  becomes 


112.  Illustrative  Problems. 

i.  What  is  the  maximum  tensile  stress  in  £/2L3  (Fig.  117)  for 
an  £-40  loading? 

The  train  should  move  on  to  the  span  at  L%  and  move  to  the 
left  as  long  as 

P2<^-  P 

32 


SEC.  IV  BRIDGES  175 

Try  wheel  3  at  L3. 


150 

163 

109 


54 

2 


1 08 
284 

392  =  total  load  on  span 

392  X  -  =  37<3°  O.K. 
32  5° 

Wheel  3  at  L3  satisfies  the  criterion,  for  when  wheel  3  is 
approaching  Z,3  the  load  in  the  panel  is  less  than  %2  °f  the 
total  load  on  the  span  and  when  wheel  3  has  passed  L3  the  load 
in  the  panel  is  greater  than  %2  of  the  total  load  on  the  span. 

16,364 

284  X  54  =  I5.336 
542  =     2,916 

240)34,616 

144-23  =  left  reaction 

230 

=  7.67  =  floor-beam  load  at  L2 

144.23  X  1 80  =  25,962 
7.67  X  240  =     1,840 


216)24,122  =  Mj 

111.7  =  maximum  tension  in 

2.  What    is    the    maximum    compressive    stress    in 
The  train  should  move  on  to  the  span  at  Z,0  and  move  to  the 
right  so  long  as 

P*<\P 
36 

Try  wheel  2  at  L2. 

Wheel  ii  is  4  ft.  on  the  span  and  the  total  load  is  172. 


176  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

Try  wheel  3  at  L2. 

Wheel  12  is  4  ft.  on  the  span  and  the  total  load  is  192. 

192X^-37^0X1 

6,708 
192  X  4  =      768 

240)7.476  »: 

31.15  =  right  reaction 

230 

=  7.67  =  floor-beam  load  at  L$ 

31.15  X  420  =  13,083 
7.67  X  270  =     2,070 

216)11,013  =  MI 

51  =  maximum  compression  in  U^L^. 

If  the  member  U^Lz  is  made  of  eye-bars  which  are  not  capable 
of  taking  compression,  it  may  be  found  necessary  to  introduce  a 
counter  LzU3  in  the  panel.  The  lever  arm  from  the  point  I  to 
this  member  is  199.7,  hence,  the  maximum  live  load  tensile 
stress  which  this  counter  would  carry  is 

^^  -  55-2 
199.7 

Since  the  chord  members  UaU*  and  LsL4  are  parallel,  the 
position  of  the  train  for  maximum  stress  in  U^L*  is  obtained 
from  the  criterion  for  shear  in  panel  3-4;  i.e.,  one-eighth  of  the 
load  on  the  span  is  placed  in  the  panel  whether  loading  from  the 
right  for  tension  or  from  the  left  for  compression. 

It  has  been  shown  that  the  criterion  for  maximum  tension  in 
UJLz  is  not  the  same  as  the  criterion  for  maximum  compres- 
sion. In  this  respect,  members  UiL%,  and  £/2Z,2  are  in  the 
same  category  as  U^Ls.  The  criterions  are  not  the  same  for  any 
two  members,  neither  are  the  criterions  for  maximum  tension 
and  compression  the  same  for  any  one  member.  The  center  of 
moments  for  Z72L3  and  Z73L3  is  at  the  same  point,  but  the  sec- 
tion is  not  the  same  for  both  members.  The  center  of  moments 
for  UiLi  and  UzLz  is  not  at  the  same  point  as  for  U%La  and  UaLs. 

113.  Problems. 

Draw  the  influence  line,  develop  the  criterion,  and  compute 


SEC.  IV 


BRIDGES 


177 


the  maximum  tensile  and  compressive  stresses  in  V\L\,  U\L^ 
UzLi,  UsLa  and  £73£4  (Fig-  117),  using  an  £-40  train.  Compute 
the  impact  stress  in  each  case,  using  the  formula  of  Article  105. 
114.  General  Criterion  for  Maximum  Stress  in  Web  Member 
of  a  Parker  Truss. — Let  M \  represent  the  algebraic  sum  of  the 
moments  of  all  the  forces  acting  on  one  side  (either  side)  of  the 
section  through  the  panel  2-3  (Fig.  118).  Since  the  stress  in 


FIG.  118. 


equals  Mj  -5-  h,  it  is  clear  that  the  stress  is  a  maximum 
when  Mt  is  a  maximum.  Let  a  equal  the  length  of  the  seg- 
ment LoLzj  b  equal  the  length  of  the  panel,  c  equal  the  length  of 
the  segment  L3L8,  and  e  equal  the  length  7L0;  then  a  +  b  +  c  =  I. 
The  influence  line  OTNSQ  for  MI  is  drawn  in  the  same  manner 
as  described  in  Article  no.  The  value  of  MT  may  be  deter- 
mined by  taking  the  sum  of  the  products  of  each  load  and 
its  corresponding  ordinate  in  the  influence  line.  The  loads  will 
be  combined  into  three  groups,  PI,  P2  and  P3,  corresponding 


178  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

to  the  three  portions  of  the  influence  line  QS,  ST  and  TO. 
If  yij  yz  and  y$  represent  respectively  the  three  ordinates  under 
the  centers  of  gravity  of  PI,  P2  and  PS,  then 
MT  =  P3;V3  ±  P2;V2  -  Piyi 

Let  the  loads  move  a  small  distance  d  to  the  left  in  such  a  way 
that  no  loads  cross  the  panel  points  L0,  L2,  £3  and  L8.  Or,  in 
other  words,  PI,  P2  and  PS  are  to  remain  constant.  The  change 
in  M  is 


Is  this  change  in  M1  a  positive  or  a  negative  quantity?     The 
answer  depends  upon  whether 


for  as  long  as  P2;y22  <P*yw  +  P\yn 

the  change  in  M  i  caused  by  a  movement  of  the  loads  to  the  left, 
will  be  a  positive  quantity.     From  similar  triangles 
yu:d::e  +  /:/ 


d(e  +  0  7  e  d(ce  -\-ae-\-  al) 

or      yn  =  -^-=  —  -,  y™  =  d-   and  ^22  =  -  ,. 

II  ol 

Hence  the  change  in  MIt  caused  by  a  movement  of  the  loads 
to  the  left,  will  be  a  positive  quantity  as  long  as 
d(ce  +  ae  +  al)         de          d(e  +  I) 


or  as  long  as  ce      ^_gp2  <eP3  +  (e  +  DP, 

Likewise,  the  change  in  M,,  caused  by  a  movement  of  the 
loads  to  the  right,  will  be  a  negative  quantity  as  long  as 


The  tensile  stress  in  UiLz  is  a  maximum  when  Mx  has  a 
maximum  positive  value,  or  when  the  right  portion  of  the  span 
is  loaded.  As  the  loads  come  onto  the  span  at  Lg  and  move  to 
the  left,  the  tension  in  U^Lz  increases  as  long  as 


SEC.  IV  BRIDGES  179 

Let  Pl  +  P2  +  P3  =  P  =  the  total  load  on  the  span. 


JL1 

and 

6 

eP>  =  eP> 

then          °e  + 

ae  +  al  + 

P«    <"'  f>  P     -1-     7  P 

b 

men 
or 

b 
(«  +  . 

?/*  p    ^~p    1     /p 

b 

-T  2  \cx*       i    vJL  \ 
P™  ^  ^P  _L  7P^ 

Hence  the  criterion  for  maximum  tensile  stress  in  Z72  Ls  is 


This  criterion  will  be  satisfied  when  the  critical  load  is  at 
LZJ  and  a  few  loads  are  in  panel  2-3.  In  the  case  of  all  ordinary 
train  loads,  this  criterion  will  be  satisfied  before  any  loads  pass 
L2,  in  which  case  PI  =  o  and  the  criterion  for  maximum  tensile 
stress  in  U-^Lz  becomes 


The  compressive  stress  in  U^L^  is  a  maximum  when  M i  has 
a  maximum  negative  value,  or  when  the  left  portion  of  the 
span  is  loaded.  As  the  loads  come  onto  the  span  at  L0  and 
move  to  the  right,  the  compression  in  U^Lz  increases  as  long  as 

ce  +  ae  +  al  r 


if 


b 
ce  +  ae 


and       _  (e  +  /)P2  =  (e  +  /)P2 

then      ce  +  ae  +  al  +  be±Up2<e       lP 


whence  P2 <[(e  +  l)P  -  IPZ] 

grion  for  maximum  < 
P£[(e  +  l)P  -  IP3] 


(a  +  b  +  e)/ 

Hence  the  criterion  for  maximum  compressive  stress  in  U^Ls  is 

b 


(a  +  b  +  e)l 


l8o  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

This  criterion  will  be  satisfied  when  the  critical  load  is  at 
L2,  and  a  few  loads  are  in  panel  2-3.  In  the  case  of  all  ordinary 
train  loads,  this  condition  will  be  satisfied  before  any  loads 
pass  L3;  in  which  case  P3  =  o  and  the  criterion  for  maximum 
compressive  stress  in  Z72Z,3  becomes 


<•> 


Criterions  (i)  and  (2)  are  easily  remembered  if  the  following 
observation  is  made.  In  computing  the  maximum  tension 
in  Z/2^3,  Mj  is  determined  from  the  truss  reaction  R0  at  LO,  and 
the  floor-beam  load  r2  at  L2,  thus 

M;  =  eR0  -  (e  +  a)r2  (3) 

If  the  two  chord  members  in  panel  2-3  were  parallel,  this 
criterion  for  maximum  tension  in  £/2L3  would  be 

(4) 


Criterion  (4)  may  be  transformed  into  criterion  (i)  by  insert- 

/> 

ing  the  fraction  -       -  in  the  coefficient  of  P.     The  numerator 
e  +  a 

and  denominator  of  this  fraction  appear  as  multipliers  in  Eq. 

(3). 

Likewise  in  computing  the  maximum  compression  in  U^Ls, 
Mz  is  determined  from  the  truss  reaction  R&  at  LS,  and  the 
floor-beam  load  rs  at  L3  thus 

Mz  =  (e  +  l)R8  -  (e  +  a  +  b)r3  (5) 

If  the  two  chord  members  in  panel  2-3  were  parallel,  the 
criterion  for  maximum  compression  in  Z72L3  would  be  criterion 
(4);  which  may  be  transformed  into  criterion  (2)  by  inserting 

e  +  / 

the  fraction  -  :  —  i  in  the  coefficient  of  P.     The  numerator 

e  +  a  +  0. 

and  denominator  of  this  fraction  appear  as  multipliers  in  Eq. 

(5)- 

Suppose  (in  Fig.  118)  there  are  eight  equal  panels.  For 
maximum  stress  in  J72L3,  the  proportion  of  total  load  on  the 
span  which  should  be  placed  in  the  panel  is  %  times  a  fraction. 
The  numerator  of  the  fraction  is  the  distance  from  the  center 


SEC.  IV  BRIDGES  l8l 

of  moments  to  the  truss  reaction,  and  the  denominator  is  the 
distance  from  the  center  of  moments  to  the  floor-beam  load. 

Thus   in   Fig.    117    the   criterion   for  maximum  tension  in 
UiLz  is 

<i  180 

>  8  ^  180  +  60 

or  •*  2>       •* 

as  given  on  page  166.     The  criterion  for  maximum  compression 
in  UiLz  is 

p  <  i       1 80  +  240  • 

>  8        180  +  90 

or  P2>  4  P 

36 

as  given  on  page  167. 

From  similar  triangles  (Fig.  118) 

ce  ce      a  (     \   j\ 

therefore         NV:c::b:a  +  c  +  - 

e 

_    .  .    al 

or 


whence  NV:NV  +  VO::b:a  +  b  +  c  H — 

NV      b     e 
or 


/ e + a 
substituting  in  criterion  (i) 


If  the  criterion  is  expressed  as  an  equation,  then 

A       I_ 

NV  ~'=  NO 

Thus  it  is  clear  that  when  the  right  portion  of  the  truss  is 
loaded  for  maximum  tensile  stress  in  t/VLs,  the  load  in  the 
panel  divided  by  the  length  NV  equals  the  total  load  on  the 
span  divided  by  the  length  NO.  Likewise  it  may  be  shown 
that  for  a  maximum  compressive  stress  in  UzLz,  the  left  portion 


182 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


of  the  truss  is  loaded  so  that  the  load  in  the  panel  divided  by 
N U  equals  the  total  load  on  the  span  divided  by  NQ. 

Compute  the  lengths  NV  and  NU  in  Fig.  117  and  show  that 
the  criterions  developed  therefrom  are  the  same  as  previously 
given. 

115.  Tension  in  a  Vertical  when  Counters  are  Used. — The 
maximum  tensile  stress  in  Z72L2  (Fig.  117)  for  an  £-40  train 
with  wheel  4  at  L2  is  68.7.  The  conditions  are  different  if  the 


FIG.  119. 

diagonal  web  members  are  designed  to  take  tensile  stresses 
only,  as  in  Fig.  119,  and  the  use  of  counters  becomes  necessary. 
The  dimensions  of  the  truss  are  the  same  as  in  Fig.  117.  We 
shall  assume  that  total  dead  load  for  the  bridge  is  2,600  Ib.  per 
linear  foot,  or  1,300  Ib.  to  be  carried  by  each  truss;  of  which  950 
Ib.  will  be  considered  as  acting  at  the  bottom  chord  and  350  Ib.  at 
the  top  chord.  The  panel  loads  are  28,500  Ib.  for  the  bottom 
chord  and  10,500  Ib.  for  the  top  chord.  The  resulting  dead 
load  stresses  for  U^Li  and  U^L^  are  shown  in  the  figure. 

Let  the  train  advance  on  the  span  at  LQ  until  wheel  7  is  at 
LQ.     The  live  load  compressive  stress  in  U^Ls  is 

2,155  xx  42Q  _ 


240 


SEC.  IV  BRIDGES  183 

and  the  impact  compressive  stress  is 

17.45  X  777  =  *S-SS 

OO  I 

The  live-load  and  impact  compressive  stresses  in  U<tLs  balance 
approximately  the  dead-load  tensile  stress,  and  the  total  stress 
in  the  member  is  zero. 

Let  the  train  advance  upon  the  span.  During  this  advance- 
ment the  counter  L<tUs  comes  into  action,  and  when  wheel  3 
arrives  at  Lv  the  live-load  tensile  stress  reaches  the  maximum 
of  55.2,  as  shown  in  Article  112.  As  the  train  advances  from 
this  position  the  tensile  stress  in  the  counter  decreases,  and 
when  wheel  16  is  4  ft.  on  the  span  the  member  ceases  to  act. 
Thus,  for  a  second  time  the  total  stress  in  L2Z73  and  in  UsLs  is 
zero,  as  the  following  computations  will  prove. 
The  reaction  at  L8  is 

_.  ....  An 
54-47 


240 
The  floor-beam  load  at  L\  is 

70 

****** 

The  floor-beam  load  at  L$  is 

rz  ,  1.785  +  230  =  6 

30 

Moments  about  /  of  the  forces  on  the  right  of  the  section 
through  panel  2-3. 

54.47  X  420  =  22,877 

2.33  X  300  =        700 
67.17  X  270  =  18,135     18,835 
216)4,042 

18.7  =  L.L. 

18.7  X  ^       14^=7. 

397       32.8  =  L.L.  +  I. 

The  live-load  and  impact  compressive  stresses  in  UzL3 
balance  (approximately)  the  dead  tensile  stress  and  the  result- 
ing stress  in  U2L3  is  zero.  Thus  when  wheel  16  is  4  ft.  on  the 
span,  the  counter  ceases  to  act  and  the  main  diagonal  is  about 


184  THEORY   OF  FRAMED    STRUCTURES  CHAP.  IV 

to  take  stress  as  the  train  advances.  This  is  evidently  the 
greatest  amount  of  train  load  which  the  truss  supports  when 
UZL3  is  inactive,  and  is  the  correct  position  of  the  train  for 
maximum  live-load  tensile  stress  in  UyLz,  which  must  balance 
the  difference  in  the  vertical  components  of  the  stresses  in 
UiUz  and  UzUz.  Instead  of  solving  by  the  method  of  joints 
we  shall  adopt  the  shorter  method  of  passing  a  section  through 
UiUz,  UzLz  and  L2L3,  and  balancing  the  moments  of  the  forces 
on  the  right  of  the  section  about  /,  as  follows: 

54.47  X  300  =  16,341 

2.33  X   180  =        420 
67.17  X  150  =  10,075     i 


120)5,846 

48.7  =  L.L. 

48.7  X  32°  =  3M  =  /. 

397       85.5  =  L.L.  +  I. 

Thus  the  maximum  live-load  and  impact  tensile  stress  in 
when  counters  are  used,  is  85.5. 

SEC.  V.    THE  BALTIMORE  TRUSS 


116.  The  Baltimore  Truss.  —  The  Pratt  truss  shown  in  Fig. 
120  has  about  the  proper  length  of  panel  for  economy,  but  the 
economic  height  of  this  truss  gives  too  steep  a  slope  for  the 
diagonal  web  members.     The  slope  of  the  diagonals  in  Fig.  121 
is  more  nearly  correct  for  an  economic  design  but  the  panels 
are  too  long.     The  Baltimore  truss  (Fig.  122)  is  an  evolution 
of  the  Pratt  truss  (Fig.  121);  for  if  an  intermediate  floor  beam 
is  placed  at  the  middle  of  each  panel  in  Fig.  121  and  supported 
by  a  sub-vertical,  M3L3  for  example,  and  the  sub-tie  M^U*  is 
added  to  prevent  bending  in   U^L^,   the  Baltimore  truss  of 
Fig.  122  is  the  result.     In  the  end  panel  the  sub-strut  MiL2  is 
used.     Sub-struts  instead  of  sub-ties  are  frequently  used  in 
the  other  panels,  as  shown  in  Fig.  92. 

117.  L4L6.  —  The  center  of  moments  for  this  member  is  at 
£/4  (Fig.  122)  and  the  section  is  passed  through  the  panel  4-5, 


SEC.  V 


BRIDGES 


and  it  is  clear  that  the  influence  line  is  the  same  as  for  L±L5 
(Fig.  120)  and  L4L6  (Fig.  121).  Hence,  the  criterion  for 
maximum  stress  is  the  same  for  all  three  members.  The  criti- 
cal wheel  is  at  L4,  and  one-third  the  total  load  on  the  span  is 
placed  on  the  segment  LoL4.  The  maximum  tensile  stress 
occurring  when  wheel  13  is  at  L4,  is  374.7. 

118.  U2U4. — The  stress  in  U<JJ*  (Fig.  121)  is  the  same  as  that 
in  L^LS,  whereas  in  Fig.  122  the  sub-tie  MsZ74  introduces  a 
slight  complication.  The  center  of  moments  is  at  L4,  the 


UR       Ug       Uio       Un 


FIG.  121. 

section  is  passed  through  the  panel  2-3;  and  it  is  important  to 
note  that  the  floor  beam  at  Lz  intervenes  between  the  section 
and  the  center  of  moments. 

As  the  load  of  one  panel  moves  from  L\i  to  LS  (not  L4  as 
previously)  the  only  force  acting  on  the  left  of  the  section  is 
the  left  reaction  R0,  and  the  moment  about  L4  is  ioa£0;  hence 
QK  =  100  ft.-lb.,  and  OT  is  the  influence  line  for  moment 
about  Z,4  when  the  load  of  i  Ib.  is  between  Li2  and  L3.  As  the 
load  of  one  panel  moves  from  L0  to  Liy  the  only  force  acting 
on  the  right  of  the  section  is  the  right  reaction  Rn;  hence,  OH 


l86  THEORY  OF  FRAMED  STRUCTURES       CHAP.  IV 

=  200  ft.-lb.  and  QS  is  the  influence  line  for  moment  about  L4 
when  the  load  of  i  Ib.  is  between  LQ  and  L2. 

TV  =  5  x  ioo  =  75 

4 

SU  =  -  X  200  =  33.3 
6 

Let  the  load  of  i  Ib.  be  in  the  panel  2-3  at  a  distance  x 
from  L3,  and  let  R0  represent  the  reaction  at  L0  and  r2  the  floor 
beam  load  at  L2;  then 

„        225  +  x       ,  # 

^o  =  —  —  -  —  and  rz  = 

300  25 

Let  M  represent  the  algebraic  sum  of  the  moments  about  L4 
of  all  the  forces  on  one  side  (either  side)  of  the  section,  then 
M  =  iooR0  —  50^2 


M  = 


-2X= 


3  3 

This  equation,  being  of  the  first  degree,  may  be  represented 
by  a  straight  line.     When  the  load  of  i  Ib.  is  at  L3,  x  =  o  and 

M  =  ^  =  75  =  TV 

o 

When  the  load  of  i  Ib.  is  at  L2,  #  =  25  and 

,,       ice  CTT 

M  =  -  =  33.3  =  SU 

o 

Since  the  influence  line  OTSQ  has  three  different  slopes,  we 
shall  let 

PI  =  the  load  on  the  segment  L0L2. 
P%  =  the  load  on  the  segment 
PS  =  the  load  on  the  segment 
The  criterion  for  maximum  moment  about  L±  of  all  the 
forces  acting  on  one  side  (either  side)  of  the  section  through 
panel  2-3,  developed  in  the  same  manner  as  explained  in  the 
preceding  sections,  is 

2Pl  +  5^2>P3 

Adding  PI  +  P2  to  both  sides  of  the  inequality,  and  letting 
Pi  +  Pz  +  PS  =  P,  we  have, 

->Pi 

o 


SEC.  V  BRIDGES  187 

The  critical  wheel  is  at  L3  and  the  position  of  the  train  for 
maximum  stress  in  U<JJ \  should  not  differ  materially  from  the 
position  of  maximum  stress  in  LJL^  previously  given. 

Try  wheel  9  at  L3. 

The  length  of  uniform  load  on  the  span  is  164  ft. 

P 

-  =  204 

3 

When  wheel  9  is  approaching  L3,  wheel  5  is  approaching  L2  and 

Pi  =  70 


When  wheel  9  has  passed  L3,  wheel  5  has  passed  L2  and 

Pi  =  90 

2P2   =IO4 

>i88 
hence  24o> 

194 

and  wheel  9  at  L3  will  not  give  a  maximum  stress  in  UzU*. 
Try  wheel  10  at  L3. 

P 

-  =  209.3 

o 

Pi  =  go 

When  wheel  10  is  approaching  L3,  P2  =  52;  and  when  wheel  10 
has  passed  L3,  P2  =  62,  hence 

>J94 

209.3> 


and  the  criterion  is  satisfied. 

Wheel  ii  at  L3  gives 

>i88 

Wheel  12  at  L3  gives 

2I8'3><255 
Wheel  13  at  Z,3  gives 


1  88  THEORY    OF   FRAMED    STRUCTURES  CHAP.  IV 

It  is  evident  that  wheels  10,  n  and  12  satisfy  the  criterion, 
and  the  algebraic  sum  of  the  moments  about  L4  of  all  the  forces 
acting  on  one  side  (either  side)  of  the  section  through  panel 
2-3  must  be  computed  for  each  condition  to  determine  the 
maximum. 

Wheel  10  at  Ls.  —  The  forces  on  the  left  of  the  section  are  the 
truss  reaction  R0,  the  loads  PI  between  L0  and  L2,  and  the 
floor-beam  load  YI  at  L&  resulting  from  the  loads  in  panel  2-3. 

*-  24*36  -3,5.99 

The  moment  of  P\,  about  L±  is 

830 

QO  X    58   =    5,220 

6,050 

The  floor-beam  load  at  Z/2  resulting  from  the  loads  in  the  panel  is 
.  =          =  33-8 


100^0  =  3J>599 

moment  of  P\  =  6,050 
=  1,664 


60)23,885 

398.1  =  compressive  stress  in 
when  wheel  10  is  at  L^. 


It  is  evident  that  these  computations  may  be  simplified  as 

follows  : 

jJS  v  f.  -.".•-- 

3)94,796 


31,599 
830 

QO  X  58  =  5,220 
832  X  2  =  1,664     7,7U 
60)23,885 
398.1 


SEC.  V  BRIDGES  189 

Wheel  ii  atLz. 

16,364 

284  X  1  80  =  51,120 
i8o2  =  32,400 

3)99,884 
33>295 


116  X  52  =  6,032 
561  X    2  =  1,122        9,309 
60)23,986 


399.8  =  compress!  ve     stress     in     UiU* 
when  wheel  n  is  at  L%. 

Wheel  12  at  L$. 

16,364 

284  X  185  =  52,540 

i852  =  34,225 

3)103,129 

34,376 
2,851 

129  X  51  =  6,579 
503  X    2  =  i,  006      10,436 
60)23,940 

399    =    compressive    stress   in    UzU* 
when  wheel  12  is  at  L3. 

Wheel  ii  at  L3  evidently  gives  the  largest  maximum  compres- 
sive stress  in  UiU*,  although  there  is  little  difference  in  the 
stress  for  the  three  positions  which  give  a  maximum. 

119.  In  Fig.  122  let  Z,oL2  =  a,  L2L3  =  ft,  LJLi  =  b  zuidLtLu 
=  c.  Let  PI  =  the  load  on  the  segment  0-2,  P2  =  the  load 
in  panel  2-3,  and  P  =  the  total  load  on  the  span.  Show  that 
the  criterion  for  maximum  stress  in  UzU*  is 

P  >  P!  +  2P2 


/  <      I  -  c 

120.  Suppose  that  the  sub-tie  M^U^  is  omitted,  and  the  sub- 
strut  L2M3  is  substituted.     Let  a,  b  and  c  represent  the  same 


THEORY   OF   FRAMED   STRUCTURES  CHAP.  IV 

lengths  as  before.  If  PI  =  the  load  on  the  segment  0-3,  P2  = 
the  load  in  panel  3-4,  and  P  =  the  total  load  on  the  span,  show 
that  the  criterion  for  maximum  stress  in  L2Z,4  is 


121.  U2M3.  —  The  influence  line  is  drawn  for  shear  in  panel 
2-3,  and  is  the  same  in  all  respects  as  the  influence  line  for  shear 
in  the  corresponding  panel  of  Fig.  120.     Hence  the  criterion  is 
the  same  in  both  cases,  or  in  other  words  one-  twelfth  of  the  total 
load  on  the  span  is  placed  in  the  panel  for  the  maximum  posi- 
tive or  negative  shear.     The  stress  in  U^Ms  (Fig.  122)  will  not 
be  the  same  as  the  stress  in  Z72Z,3  (Fig.  120),  although  the 
maximum  shear  in  both  cases  is  the  same;  for  the  two  members 
do  not  have  the  same  slope.     The  maximum  positive  shear 
(wheel  3  at  Lz)  is  222.9;  the  maximum  negative  shear  (wheel  2 
at  L2)  is  13.2. 

122.  MsL4.  —  The  influence  line  will  be  drawn  for  the  vertical 
component  of  the  stress.     When  the  load  of  i  Ib.  is  at  any  point 
on  the  span  except  between  L2  and  L±,  there  is  no  stress  in  M  sLa 
or  MzU±  and  the  stress  in  M$Li  is  the  same  as  the  stress  in 
U<iMz.     Consequently  the  influence  line  for  M^Li  will  be  the 
same  as  for  U*M*,  except  when  the  load  of  one  pound  is  between 
L2  and  L4. 

TV  =  + 


Let  the  load  of  i  Ib.  be  in  panel  3-4  at  the  distance  x  from 
L4.  Let  RQ  represent  the  truss  reaction  at  L0;  r3,  the  floor-beam 
load  at  La;  and  r2  the  floor-beam  load  at  L2;  then 

200  +  X  X 

RQ  =  -         —">rz  =  —  and  r2  =  o 
300  25 

The  web  members  meeting  at  Mz  are  shown  in  Fig.  123.  Let 
V  represent  the  vertical  component  of  the  stress  in  MsLt,  for 
which  we  are  drawing  the  influence  line.  The  vertical  com- 
ponent of  the  stress  in  UzM  3  is  the  shear  in  panel  2-3,  which  is 
RQ.  Balance  the  moments  of  all  the  forces  about  Ut,  thereby 


SEC.  V 


BRIDGES 


IQI 


•M 


"         60' 


FIG.  122. 


I  Q2  THEORY   OF   FRAMED   STRUCTURES  CHAP.  IV 

eliminating  all  the  forces  except  the  force  at  La  and  the  vertical 
component  at  Z/4,  whence  the  vertical  component  at  U*  is 

257-3       i 
-  =  -  TZ 

50  2 

Balance  the  vertical  components 

V  +  rz  =  Ro  +  -  r, 

2 

T/  r>  I  200  +  X  X 

V    =   KQ  --  TZ   =  -  -    -- 

2  3°°  50 

v  _  200  -  5* 

300 

Now  let  the  load  of  i  Ib.  be  in  panel  2-3  at  a  distance  x  from 
L4,  then 

D  200  +  X  50  —  #  tf  —   25 

RQ  =  ~         —  »  7-3  =  -  -  7  and  ^2  =  -  - 

300  25  25 

The  vertical  component  of  the  stress  in  UiM^  (Fig.  124)  equals 
the  shear  in  panel  2-3,  which  is  R0  —  r2,  and 

V  +  rz  =  RQ  -  r2  +  -  r3 

2 

7  _  200  -    * 


300 

Hence,  when  the  load  is  at  a  distance  x  from  £4,  whether  in 
panel  3-4  or  panel  2-3,  the  vertical  component  of  the  stress  in 
M  sZ/4  is  given  by  the  equation 

v  =  200  -  $x 

300 

which  may  be  represented  by  a  straight  line.  When  the  load 
of  i  Ib.  is  1,4,  x  =  o  and 

V  =  +  -  =  TV 
3 

When  the  load  of  i  Ib.  is  at  L^  x  =  50  and 

V  =  -  §  =  US 

6 

Hence  the  straight  line  TS  is  the  influence  line  for  the  vertical 
component  of  the  stress  in  M^L^  when  the  load  of  i  Ib.  is  between 
£2  and  L4. 

It  is  clear  that  the  influence  line  OTNSQ  is  also  the  influence 


SEC.  V  BRIDGES  193 

line  for  shear  in  panel  2-4  (Fig.  121)  ;  or  in  other  words,  it  is  an 
influence  line  for  the  vertical  component  of  the  stress  in  UzL*. 
Obviously  then,  the  criterion  for  a  maximum  stress  in  M  $Li 
(Fig.  122)  is  the  same  as  for  maximum  stress  in  U2L^  (Fig.  121), 
likewise  the  maximum  stress  is  the  same  in  both  members. 
In  order  to  show  this  more  clearly,  we  shall  compute  the  maxi- 
mum tensile  stress  in  UiLi  (Fig.  121). 


Try  wheel  5  at  L±. 

23  16,364 
200                         284  X  114  =  32>376 

223                         ii42  =  12,996 

109  300)61,736 

114  205.79  =  ^o 

**?  =  16.6    =r> 

5° 

228  189.19  =  shear  in  panel  2-4 

284 


6)512 


70 


85.3  >    ~O.K. 
*  <  go 

189.19  X  — —  =  295.5  =  maximum  tensile  stress  in 

If  a  section  is  passed  through  panel  3-4  (Fig.  122),  it  is  evident 
that  the  vertical  forces  acting  on  the  left  of  the  section  must 
balance.  The  vertical  component  of  Mali*  acting  upward  on 
the  left  portion  of  the  truss  is  %rs.  The  forces  acting  upward 
on  the  left  of  the  section  are  R0  and  ^r3,  and  they  are  balanced 
by  the  floor-beam  loads  on  the  left  of  the  section  and  the 
vertical  component  V  in  M$  L± ;  hence 

#o  +  -  r3  =  r2  +  r3  +  V 

2 

When  wheel  5  is  at  £4,  wheel  i  is  in  panel  3-4 

rz  =  o 
f3  =  33-2 

RQ   =    205.79 

hence  V  =  R0 r3  =  189.19 


IQ4  THEORY   OF   FRAMED    STRUCTURES  CHAP.  IV 


Thus  it  is  clear  that  the  tensile  stress  in  M^L*  (Fig.  122)  is  the 
same  as  in  V^L^  (Fig.  121). 

124.  U4L4.  —  The  influence  line  for  stress  in  U  *Li,  when  the 
sub-members  M  zLs  and  MsU*  are  omitted,  is  OTUQ.  The 
presence  of  the  sub-members  alters  the  influence  line  between 
L2  and  £4.  When  the  load  of  i  Ib.  is  between  Z,2  and  L4 
the  sub-members  take  stress,  and  the  stress  in  Z74L4  when  the 
load  of  i  Ib.  is  at  L3  is  J^  Ib.  compressive.  It  may  be  shown 
by  the  method  used  on  previous  occasions  that  the  influence 


FIG.  124. 

line  between  panel  points  is  a  straight  line,  therefore  OTUSVQ 
is  the  influence  line  for  stress  in  £74L4. 

Let  PI  =  the  load  on 
Pz  =  the  load  on 
Pz  =  the  load  on  L3L4 
P4  =  the  load  on  L±Lz 
PS  =  the  load  on  L^L^ 
P  =  total  load  on  the  span 

The  criterions,  when  developed  as  in  preceding  cases,  will 
show  that  the  compressive  stress  in  £74L4  will  be  increasing  as 
the  loads  move  to  the  left  as  long  as 


add 


or  as  long  as       P  +  6P3  >  6P2  +  i2P4 


P 
6 
Two  conditions  of  loading  should  be  investigated.     The  train 


or  as  long  as       -^-\-    P3>  PI  -f 


SEC.  V  BRIDGES  195 

may  advance  from  the  right  until  a  few  loads  pass  L6,  in  which 
case  PI,  P%  and  P$  are  zero,  and  the  criterion  becomes 


which  is  obviously  the  same  as  the  criterion  for  UiM$.  Or  the 
train  may  advance  until  the  compressive  area  of  the  influence 
line  at  S  is  covered.  Since  the  compressive  area  at  5  is 
smaller  than  the  tensile  area  at  U,  it  is  doubtful  if  the  second 
position  of  the  train  will  give  a  larger  compressive  stress  than 
the  first;  although  if  large  wheels  are  near  L3  and  smaller 
wheels  near  L4,  the  difference  in  areas  will  not  be  so  significant. 
For  the  first  position  of  this  train,  wheel  3  at  L6  satisfies  the 
criterion 

£ft 

and  the  maximum  compressive  stress  in  Z74L4  is  140.94. 
For  the  second  position  of  this  train  the  criterion  is 

\  +  P£P*  +  2P, 

and  the  critical  wheel  will  be  at  either  L3  or  L&. 
Try  wheel  3  at  L3 

WHEEL  3  APPROACHING  L^  WHEEL  3  PASSING  La 

Pi  =     o  Pi  =       o 

P2  =30  P2  =     50 

P3  =86  P3  =    66 

P4  =36  P4  =    36 

PS  =  39°  p6  =  390 

P    =  542  P    =  542 

P  P 

-    =     90.3  P2  =     30  -    ==     90.3       P2  =   50 

P3  =     86  2P4  =     72  P3  =    66         2P4  =     72_ 

176.3      >  102  IS6.3  >  122 

In  either  case  the  compressive  stress  in  £74L4  is  increasing. 
Try  wheel  4  at  L3 

WHEEL  4  APPROACHING  X3  WHEEL  4  PASSING  L3 

P  P 

-  =    92  P2  =    50  -    =    92  P2  =  70 

6  o 

P3  =  66     2P4  =  112       P3  =  _S9     2P4  =  86 
158     <   162  151     <     156 


Ip  THEORY   OF   FRAMED   STRUCTURES  CHAP.  IV 

In  either  case  the  compressive  stress  in  £74L4  is  decreasing. 
Since  the  stress  is  increasing  when  wheel  3  passes  £3,  and  is 
decreasing  when  wheel  4  arrives  at  L3;  there  must  be  an  inter- 
mediate position  for  which  the  stress  is  a  maximum.  Try 
wheel  ii  at  Z,5 

WHEEL  1 1  APPROACHING  Z,6  WHEEL  1 1  PASSING  L6  .  ,? 

p 
-=90.7       P2  =     50  P6=90.7  P2  =     50 

P3  =  _66_      2P4  =     72  P3  = 66  2P4  =  112^ 

156.7      >  122  156.7  <  l62 

Wheel  ii  at  L$  satisfies  the  criterion  for  maximum  compres- 
sive stress.  Pass  a  section  cutting  the  members  UzU*,  M3£/4, 
Z74L4  and  L^L5  and  consider  the  forces  acting  on  the  left  ot 
the  section.  The  forces  acting  upward  are  the  truss  reaction 
R0  =  233.95,  and  tne  vertical  component  of  the  stress  in 
MsUi  =  \r$  =  36.54.  The  forces  acting  downward  are  the 
sum  of  the  wheel  loads  on  the  left  of  Z,4,  which  is  116,  and  the 
floor-beam  load  at  L4  due  to  the  loads  in  panel  4-5,  which  is 
22.44.  Hence  the  compressive  stress  in  Z74L4  is  132.05.  This 
stress  is  somewhat  less  than  that  previously  determined  for 
wheel  3  at  L5. 

For  the  maximum  tensile  stress  in  Z74L4,  the  train  approaches 
from  the  left  and  it  is  obvious  from  the  influence  line  that  the 
train  will  advance  until  the  large  wheels  of  the  front  engine  are 
atZ4. 
Try  wheel  3  at  Z4 

WHEEL  3  APPROACHING  L4  WHEEL  3  PASSING  Z4 
PI  =  140  PI  =  140 

Pz  =36  P2  =    36 

P3  =86  Pz  =    66 

P*  =30  P*  =    50 

P5  =      o  P6  =      o 

P    =  292  P    =  292 

--    48.7      P2  =  36  ^=    48-7      P*=    36 

P3  =    86        2P4  =60  P3  =     66        2P4  =  100 

134-7    >  96  124.7    <  136 

Wheel  3  at  L4  satisfies  the  criterion  for  maximum  tensile 
stress  in  Z74Z,4.  Consider  the  forces  acting  on  the  right  of  the 
section  through  UzU*,  M3U^  £74L4  and  LJL6.  The  forces 


SEC.  VI  BRIDGES  IQ7 

acting  upward  are  the  truss  reaction  R^  =  58.39  and  the  floor 
beam  load  r$  =  9.2.  The  vertical  component  of  MsUi  =  ^3 
=  27.68  acts  downward.  Hence  the  tensile  stress  in  UiL± 
when  wheel  3  is  at  L4,  is  21.51. 

It  may  be  interesting  to  investigate  the  maximum  tensile 
stress  in  U^L*  when  the  train,  approaching  from  the  left,  has 
advanced  until  only  a  few  loads  are  on  panel  2-3;  in  which 
case  PZ,  PI  and  P$  are  zero  and  the  criterion  reduces  to 


Wheel  2  at  1,2  satisfies  this  criterion.  R^  =  16.45,  r*  ~  3-2 
the  maximum  tensile  stress  in  UJL,  for  this  position  is  16.45  ~~~ 
1.6  =  14.85;  which  is  less  than  in  the  preceding  case,  when 
wheel  3  has  advanced  to  L±. 

125.  UJui.  —  When  the  i  Ib.  load  is  at  LI,  the  vertical  com- 
ponent of  the  stress  in  M  iL2  causes  a  tensile  stress  of  J^  Ib.  in 
UzLz.     Hence  the  influence  line  is  a  triangle  with  the  apex  at 
L2    and    a   base  line  from  L0  to  L3.     For   maximum  tensile 
stress  the  train  comes  on  the  span  at  L0  and  one-  third  of  the 
total  load  is  in  panel  2-3.     The  stress  equals  the  algebraic  sum 
of  the  moments  about  L0  of  the  truss  reaction  at  L\i  and  the 
floor-beam  load  at  L3,  divided  by  the  distance  LoL2. 

In  case  the  main  diagonals  are  designed  to  carry  a  tensile 
stress  only,  and  it  is  found  that  counters  are  required  (for 
example  in  panel  4-6)  the  member  L±M  5  is  added.  The  mem- 
bers LtMz  and  M&UQ  then  become  the  main  diagonals.  The 
diagonal  U^M5  becomes  a  sub-tie  and  the  vertical  component 
of  its  stress  is  one-half  the  stress  in  M^L5,  and  the  member 
JW6L6  is  not  acting  at  all. 

SEC.  VI.    THE  PENNSYLVANIA  TRUSS 

126.  Pennsylvania  Truss.  —  The  influence  lines  for  several 
members    of    a    Pennsylvania    truss  are  shown  in  Fig.   125. 
This  truss  is  an  evolution  of  the  Baltimore  truss,  and  bears  the 
same  relation  to  it  that  exists  between  the  Parker  and  the  Pratt 
trusses. 


i98 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


127.  U2U4. — The  influence  line,  the  criterion  and  position 
of  a  train  for  maximum  stress  in  this  member  are  the  same  as 
for  the  corresponding  member  of  the  Baltimore  truss  in  Fig. 


-I28)?5=300[ -,-> 


FIG.  125. 

122.     The  stresses  differ  only  because  the  lever  arms  from  L* 
to  the  members  are  different  in  the  two  trusses. 

128.  I^Ma. — The  section  is  passed  through  panel  2-3,  and 
the  chord  members  intersect  at  7.  In  the  Baltimore  truss, 
one-twelfth  of  the  total  load  on  the  span  is  placed  in  panel  2-3 
for  maximum  tensile  or  compressive  stress.  In  the  Pennsly- 


SEC.  VI  BRIDGES  1  99 

vania  truss,  the  train  covers  the  right  portion  of  the  truss, 
with  the  critical  wheel  at  L%  for  a  maximum  tensile  stress,  and 
the  moment  about  /  of  all  the  forces  on  the  left  of  the  section  is 

Mz  =  i37-5^o  -  I87-5/-2 

Let  PI  represent  the  load  in  panel  2-3,  and  P  the  total  load 
on  the  span;  then  the  criterion,  developed  as  in  Article  in  or 
114,  is 

P8<JL  I1715p 

>I2    187.5 

or  P2>o.o6iP 

The  criterion  may  also  be  determined  by  the  ratio, 

NV         14.6 

=  —  -  -  =  0.061 

NO       239.6 

For  a  maximum  compressive  stress  the  left  portion  of  the  truss 
is  loaded  with  the  critical  wheel  at  1/2,  and 

MI  =  437-5^12  —  212.573 

and    the    criterion   is  P2  ^—  ^^ 

12  212.5 

or  P2>o.i72P 

This  criterion  may  also  be  determined  by  the  ratio, 
NU       10.4 


129.  MSL4.  —  As  in  the  Baltimore  truss,  the  stress  in  this 
member  is  the  same  as  the  stress  in  UiL±  when  the  sub-members 
MsLz  and  MzU*  are  removed.  This  is  clearly  shown  by  the 
influence  line.  With  the  sub-members  omitted,  the  maximum 
tensile  stress  in  UiL±  occurs  when  the  right  portion  of  the  truss 
is  loaded  and  the  critical  wheel  is  at  L4,  and 
Ml  =  137-5^0  -  187.57-2 

Let  P2  represent  the  load  on  L^L^  and  P  the  total  load;  then 
the  criterion  is 

P      -  T   I37-5P 
-  6  IS^ 


Or  P2    =   O.I22P 

NV  27.8 
^-—  =  -  - 
NO  227.8 


NV         27.8 
Likewise  -—  =  -      -  =  0.122 


200  THEORY   OF   FRAMED   STRUCTURES  CHAP.  IV 


For  a  maximum  compressive  stress  in  U2L^  the  left  portion  of 
the  truss  is  loaded  with  the  critical  wheel  at  L2,  and 

Mj  =  437-5^12  -  237.57-4 
Hence  the  criterion  is         P2<- 


237.5 
or  P2<o.3o7P 


130.  U4L4.  —  For  maximum  compression,  the  right  portion 
of  the  truss  is  loaded  with  the  critical  wheel  at  L5  and  the 
criterion  is 


12  237-5 

or  P2<o.o48P 


Likewise  =  --  =  0.048 


The  criterion  for  maximum  tensile  stress  requires  a  special 
treatment  similar  to  that  given  for  the  corresponding  case  in 
the  Baltimore  truss.  Since  the  compressive  area  ot  the  influ- 
ence line  at  L  3  is  comparatively  small,  it  is  evident  that  the 
train  in  approaching  from  the  left  should  extend  into  panel  4-5, 
with  wheel  2  or  3  at  L±. 

Counters  are  treated  in  a  similar  manner  as  outlined  for  the 
Baltimore  truss. 

SEC.  VII.     EQUIVALENT  UNIFORM  LOADS 

131.  In  the  previous  sections  of  this  chapter  the  shear  and 
moment  at  a  given  section  in  a  beam  or  girder,  also  the  stress  in 
a  given  member  of  a  truss  caused  by  a  train  load,  were  com- 
puted by  means  of  a  moment  table  Fig.  (100)  .     If  a  uniform  live 
load  were  used,  the  computations  would  be  greatly  simplified; 
as  will  be  seen  from  the  following  examples. 

132.  Examples.  —  Consider  a  uniform  live  load  of  3,000  Ib.  per 
linear  foot  in  connection  with  the  beam  of  Fig.  97.     The  influ- 
ence line  shows  that  the  uniform  live  load  will  cover  the  seg- 
ment BC  for  a  maximum  positive  shear  at  C;  and  the  segment 
AC  for  a  maximum  negative  shear  at  C.    Let  AC  =  10  ft.,  and 


SEC.  VII  BRIDGES  201 

BC  =  15  ft.;  then  TN  =  +0.6  and  NS  =  -0.4.  When  the 
live  load  extends  from  B  to  C,  the  shear  at  C  equals  the  reaction 
at  A  ;  or 

X  15  X  15  _       ,  coo  lh 
-  13,500  Ib. 


The  shear  may  also  be  obtained  by  taking  the  product  of  the 
area  OTN  and  the  intensity  of  the  uniform  load  per  foot,  thus 

pe  =  °'6  X  IS  X  3,000  =  I3,5oo  Ib. 

2 

Similarly  the  maximum  negative  shear  at  C,  occurring  when 
the  live  load  extends  from  A  to  C,  is 

Fc  =  -°-4  X-L°  X  3,000  =  -6,000  Ib. 

2 

In  Fig.  98  the  influence  line  shows  that  a  uniform  live-load 
covers  the  whole  span  for  a  maximum  bending  moment  at  C. 
For  a  uniform  live  load  of  2,000  Ib.  per  linear  foot  the  bending 
moment  at  C  is 


Mc  =  X  10  X  15  =  150,000  ft.-lb. 

2 

The  bending  moment  may  also  be  determined  by  taking  the 
product  of  the  area  QNO  and  the  intensity  of  the  uniform  load 
per  foot,  thus 

Mc  =  ^-*-?-5  X  2,000  =  150,000  ft.-lb. 

2 

In  Fig.  108  the  influence  line  shows  that  a  uniform  live  load 
extends  from  O  to  N  for  a  maximum  positive  shear  in  panel  1-2. 
NV  =  16.67  ft.  Let  the  intensity  of  the  uniform  live  load  be 
3,600  Ib.  per  linear  foot,  then, 

3,600  X66.6f 


2   X    100 

2 


3,600  X  16.672 
n  =  —  —  -  =  20,000  Ib. 

2  X  25 

and  the  shear  in  panel  1-2  is 

F  =  RQ  —  r\  =  60,000  Ib. 
The  shear  in  the  panel   may  also  be  found  by  taking  the 


202  THEORY  OF  FRAMED  STRUCTURES       CHAP.  IV 

product  of  the  area  NTO  and  the  intensity  of  the  uniform  load 
per  foot,  thus 

F  =  66-67_Xo.S  x 


133.  General  Considerations.—  In  Article  103  the  maximum 
positive  shear  in  panel  1-2  of  the  truss  in  Fig.  109  was  found  to 
be  47,760  lb.  for  an  £-40  train.  Let  q  represent  the  intensity 
of  the  equivalent  uniform  live  load  which  will  cause  the  same 
shear,  then 

q  X  area  NTO  =  47,760 

area  NTO  =  66'67  ><°-5  =  l6  6y 

q  =  ^^  =  2,866  lb.  per  linear  foot 
10.07 

Hence  it  is  clear  that  a  uniform  live  load  of  2,866  lb.  per 
linear  foot  for  one  truss  will  cause  the  same  maximum  positive 
shear  in  panel  1-2  of  the  truss  in  Fig.  109,  as  an  £-40  train. 

The  maximum  negative  shear  in  panel  1-2  for  an  £-40  train 
was  found  to  be  14,230  lb.  Let  q  represent  the  intensity  of  the 
equivalent  uniform  live  load  which  will  cause  the  same  negative 
shear,  then 

q  X  area  QSN  =  —  14,230 
area  QSN  =  -  4.167 

q  =  3415  lb.  per  linear  foot 

In  Article  98  the  maximum  moment  for  the  stress  in  UiUz 
(Fig.  106)  for  an  £-40  train  was  found  to  be  3,219,000  ft.-lb. 
The  area  of  the  influence  line  (shown  in  Fig.  105)  is  1,250; 
hence  the  intensity  of  the  equivalent  uniform  load  for  this 
member  is 

q  =  —  —  —  —  =  2,575  Hx  per  linear  foot 
1,250 

In  Article  112  the  maximum  value  for  Mz  for  the  member 
Z7  2£s  (Fig.  1  1  7)  was  found  to  be  24,  1  2  2,  ooo  ft.-lb.  The  area  of  the 
influence  line  NTO  is  9,309;  hence  the  intensity  of  the  equivalent 
uniform  load  for  maximum  tension  in  this  member  is 

24,122,000 

q  —  •  —        -  =  2,591  lb.  per  linear  foot. 
9,309 


SEC.  VII 


BRIDGES 


203 


From  a  consideration  of  the  examples  just  given,  it  is  apparent 
that  the  stresses  in  a  truss  might  be  very  quickly  computed  if  a 
uniform  load  could  be  substituted  for  the  £-40  train  load.  The 
only  hindrance  to  this  substitution  lies  in  the  fact  that  the 
equivalent  uniform  load  is  not  the  same  for  all  members  of  a 


truss.  In  fact,  the  equivalent  uniform  load  for  the  tensile 
stress  in  a  web  member  is  not  the  same  as  for  the  compressive 
stress  in  the  same  member,  as  was  clearly  shown. 

134.  Triangular  Influence  Diagrams. — Nearly  all  the  influ- 
ence line  diagrams  which  have  been  considered  in  this  chapter 
are  triangular  as  shown  in  Fig.  126.  It  is  clear  that  the  tri- 
angle ABC,  in  which  l\  —  20  ft.,  and  /2  =  80  ft.,  may  represent 


204 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


the  influence  line  for  determining  the  following  quantities  for 
an  £-40  train  by  varying  the  altitude  h. 

Let  s  =  maximum  positive  shear  in  panel  1-2,  truss  (a) 
t  =  maximum  tension  in  Z7iL2  truss  (a) 
u  =  maximum  stress  in  L§U\  truss  (b) 
v  =  maximum  shear  in  panel  o-i  truss  (b) 
w  =  maximum  stress  in  LoLz  truss  (b) 
x  =  maximum  bending  moment  at  £,  beam  (c) 
y  =  maximum  pier  reaction  at  F,  span  (d) 

0.6  Ib.  for  s 
0.76  Ib.  for  / 
0.8    Ib.  for  u 
then  h  =   {      0.96  Ib.  for  v 

0.53  Ib.  for  w 

1 6   ft.-lb.  for  x 

i.o    Ib.  for  y 

Since  li  and  /2  are  constants  in  all  these  cases,  the  criterion 
for  a  maximum  for  all  quantities  (s  to  y)  is  the  same.  This 
criterion  places  wheel  4  on  the  vertical  through  D,  and  by  the 
use  of  the  moment  table,  the  value  of  each  quantity  may  be 
found  as  given  below: 

s  =  81,982.5  Ib. 

/  =  103,844  Ib. 

u  =  109,310  Ib. 

v  =  131,172  Ib. 

w  =  72,417  Ib. 

x  =  2,186,200  ft.-lb 

y  =  136,637.5  Ib. 

Instead  of  using  the  moment  table  these  quantities  could 
have  been  determined  by  taking  the  sum  of  the  products  of 
each  wheel  load  and  its  corresponding  ordinate  in  the  influence 
line  diagram,  when  the  proper  value  for  h  is  taken.  Hence  it  is 
seen  that  the  sum  of  the  products  in  each  case  is  proportional 
to  the  corresponding  value  of  h,  as  an  inspection  will  prove;  thus 
for  example,  in  the  case  of  5  and  u, 

81,982.5:0.6:  -.109,310:0.8 


SEC.  VII  BRIDGES  205 

Since  /i  and  h  are  constants,  it  is  clear  that  the  area  of  the 
influence  line  diagram  ABC  for  each  quantity  is  also  propor- 
tional to  the  corresponding  value  of  h\  and  from  this  it  follows 
that  each  quantity  (s  to  y)  is  proportional  to  the  area  of  its 
influence  line  diagram;  or  in  other  words,  the  ratio  between 
each  quantity  and  its  influence  line  area  is  constant.  This 
constant  ratio  is  the  intensity  of  the  equivalent  uniform  load 
which  corresponds  to  any  triangular  influence  line  diagram  for 
which  /i  =  20  ft.,  and  1%  =  80  ft.  Let  q  represent  the  intensity 
of  the  equivalent  uniform  load  corresponding  to  an  £-40 
train  load.  From  the  previous  discussion  it  is  clear  that  q  is  a 
function  of  /i  and  h  and  independent  of  h. 

135.  Table  of  Equivalent  Uniform  Loads. — Table  I1  gives  the 
values  of  q,  the  equivalent  uniform  load  per  linear  foot  per  rail 
for  an  £-40  loading,  for  various  lengths  of  /i  and  /2  in  multiples 
oi  5  ft.  The  values  of  q  will  be  found  correct  in  most  instances 
to  the  third  significant  figure,  and  are  thus  accurate  to  within 
i  per  cent.  The  corresponding  values  of  q  for  any  other 
Cooper's  standard  loading  are  directly  proportional.  The 
following  example  will  be  given  to  illustrate  the  use  of  the 
table.  In  Article  112  the  maximum  tensile  stress  in  U<iL-6 
(Fig.  117)  was  found  to  be  111.7.  The  influence  line  NTO  is 
drawn  for  M/;  NV  =  15.5  =  /i;  VO  =  150  =  /2;  and  the  area 
is  9,309.  From  Table  I,  q  =  2. 5 90  expressed  in  i,ooolb. -units. 
Mz  =  9,309  X  2.59  =  24,110 

TT  ,         24,110 
UzLs  =  — -  ——  =  in. 6 
216 

This  gives  a  reasonably  close  check. 

For  the  maximum  compressive  stress  the  area  NQS  is  3,911; 
NU  =  14.5  =  /i;  UQ  =  60  =  h\  henceg  =  2. 8 20  and  the  maxi- 
mum compressive  stress  is 

3,911  X  2.82  = 
216 

1This  table  is  taken  from  p.  112,  ist  ed.,  of  "Live-load  Stresses  in  Railway 
Bridges,"  by  GEORGE  E.  BEGGS,  published  by  John  Wiley  &  Sons.  Professor 
Beggs  has  kindly  granted  permission  to  reproduce  this  table.  There  are  many 
other  tables  in  this  book  which  will  be  found  very  useful  for  ready  reference  to 
the  designing  engineer. 


206 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  IV 


TABLE  I. — EQUIVALENT  UNIFORM  LOADS  FOR  COOPER'S  £-40  LOADING  VALUES 
IN  POUNDS  PER  LINEAR  FOOT  PER  RAIL 


Longer 
segment  It 

Shorter  segment  h 

0 

5 

10 

IS 

20 

25 

30 

35 

40 

45 

50 

5S 

250 

2,500 

2,450 

2,430 

2,410 

2,380 

2,370 

2,350 

2,330 

2,310 

2,300 

2,290 

2  ,  270 

225 

2,550 

2,500 

2,460 

2,450 

2,430 

2,400 

2,380 

2  ,360 

2,340 

2,320 

2,310 

2,300 

200 

2,610 

2,540 

2,500 

2,490 

2,460 

2,440 

2,420 

2,390 

2,370 

2,350 

2,340 

2,320 

175 

2,680 

2,610 

2,550 

2,540 

2,510 

2,490 

2,460 

2,420 

2,400 

2,380 

2,360 

2,340 

160 

2,730 

2,630 

2,590 

2,570 

2  ,  540 

2,510 

2,480 

2,450 

2,420 

2,400 

2,380 

2,370 

ISO 

2,760 

2,670 

2,620 

2,590 

2,570 

2,540 

2,500 

2,460 

2,430 

2,420 

2,400 

2,380 

140 

2,800 

2,700 

2,650 

2,620 

2,580 

2,560 

2,520 

2,490 

2,450 

2,430 

2,420 

2,400 

130 

2,850 

2,740 

2,670 

2,650 

2,610 

2,580 

2,540 

2,510 

2,470 

2,450 

2,430 

2,420 

120 

2,900 

2,770 

2,710 

2,680 

2,640 

2,610 

2,560 

2,530 

2,490 

2,460 

2,450 

2,430 

no 

2,940 

2,810 

2,740 

2,710 

2,660 

2,630 

2,580 

2,550 

2,500 

2,490 

2,460 

2,460 

IOO 

3,000 

2,850 

2,780 

2,740 

2,690 

2,660 

2,610 

2,570 

2  ,  530 

2,510 

2,500 

2,480 

95 

3,020 

2,880 

2,800 

2,760 

2,700 

2,670 

2,620 

2,580 

2,560 

2,540 

2,520 

2,500 

90 

3,050 

2,890 

2,810 

2,770 

2,720 

2,680 

2,630 

2,620 

2,590 

2,570 

2,550 

2,540 

85 

3,080 

2,920 

2,820 

2,780 

2,730 

2,700 

2,640 

2,640 

2,620 

2,580 

2,570 

2,550 

80 

3,  no 

2,920 

2,840 

2,790 

2,740 

2,710 

2,670 

2,660 

2,620 

2,610 

2,580 

2,570 

75 

3,140 

2,940 

2,860 

2,800 

2,740 

2,700 

2,670 

2,660 

2,640 

2,620 

2,600 

2,580 

70 

3,i6o 

2,940 

2,870 

2,810 

2,750 

2,700 

2,670 

2,660 

2,650 

2,620 

2,600 

2,580 

65 

3,190 

2,960 

2,870:2,810 

2,760 

2,700 

2,670 

2,660 

2,650 

2,620 

2,600 

2,580 

60 

3,270 

3,020 

2,880  2,820 

2,750 

2,700 

2,660 

2,640 

2,630 

2,610 

2,590 

2,580 

55 

3,370 

3,090 

2,930  2,840 

2,760 

2,700 

2,660 

2,650 

2,620 

2,600 

2,560 

2,550 

50 

3,490 

3,i8o 

3,ooo 

2,910 

2,800 

2,740 

2,700 

2,670 

2,630 

2,600 

2,580 

45 

3,630 

3,260 

3,080 

2,980 

2,870  2,780 

2,740 

2,710 

2,670 

2,640 

40 

3,770 

3,350 

3,  1  8o|  3,  060 

2,930 

2,840 

2,?80 

2,740 

2,700 

35 

3,96o 

3,450 

3,260  3,  120 

3,010 

2,900 

2,840 

2,790 

30 

4,200 

3,6io 

3,38o 

3,200 

3,060 

2,960 

2,880 

25 

4,540 

3,770 

3,520  3,320 

3,150 

3,020 

20 

S.ooo 

4,000 

3,7303,450 

3,280 

15 

5,336 

4,000 

4,000 

3,650 

10 

6,000 

4,000 

4,000 

5 

8,000 

4,000 

SEC.  VII 


BRIDGES 


207 


TABLE  I. — EQUIVALENT  UNIFORM  LOADS  FOR  COOPER'S  £-40  LOADING  VALUES 
IN  POUNDS  PER  LINEAR  FOOT  PER  RAIL 


Longer 
seg- 
ment 
It 

Shorter  segment  h 

60 

65 

70 

75 

80 

85 

90 

95 

IOO 

no 

120 

130 

140 

250 

2,260 

2,260 

2,250 

2,250 

2,240 

2,230 

2,220 

2,22O 

2.2IO 

2,200 

2,  1  8O 

2,  160 

2,  I4O 

225 

2,290 

2,280 

2  ,270 

2,270 

2,260 

2,260 

2  ,25O 

2,240 

2,220 

2,220 

2,180 

2,  I7O 

2  ,  ISO 

200 

2,310 

2,300 

2,290 

2,290 

2,280 

2  ,28O 

2,  27O 

2,260 

2,25O 

2,230 

2,200 

2,180 

2,160 

175 

2,340 

2,320 

2,32O 

2,320 

2,310 

2  ,3OO 

2,  29O 

2,280 

2,27O 

2,240 

2,2IO 

2,200 

2,180 

160 

2,350 

2,340 

2,340 

2,340 

2,330 

2,320 

2,3IO 

2,300 

2,280 

2,260 

2,23O 

2.2IO 

2,180 

150 

2,370 

2,350 

2,360 

2,350 

2,340 

2,340 

2,330 

2,300 

2,30O 

2,270 

2,240 

2,220 

2,190 

140 

2,380 

2,380 

2,370 

2,360 

2,360 

2,350 

2,340 

2,320 

2,310 

2,280 

2,  25O 

2,220 

2,200 

130 

2,400 

2,390 

2,390 

2,380 

2,380 

2,370 

2,350 

2,340 

2,330 

2,290 

2,260 

2,230 

I2O 

2,420 

2,410 

2,410 

2,400 

2,400 

2,370 

2,370 

2,350 

2,340 

2,300 

2,  27O 

no 

2,440 

2,420 

2,420 

2,420 

2,420 

2,400 

2,390 

2,380 

2,350 

2,320 

IOO 

2,460 

2,460 

2,450 

2,440 

2,440 

2,420 

2,410 

2,390 

2,380 

95 

2,500 

2,480 

2,460 

2,460 

2,450 

2,440 

2,420 

2  ,40O 

90 

2,510 

2,500 

2,480 

2,460 

2,460 

2,450 

2,430 

85 

2,530 

2,510 

2,500 

2,490 

2,470 

2,460 

80 

2,550 

2,540 

2,520 

2,500 

2,490 

75 

2,560 

2,540 

2,530 

2,510 

70 

2,560 

2,540 

2,530 

65 

2,560 

2,540 

60 

2,550 

1               1 

CHAPTER  V 
DEFLECTION  OF  BEAMS 

SEC.  I.    THE  AREA-MOMENT  METHOD 

136.  General  Considerations. — Whenever  vertical  loads  are 
applied  between  the  supports  of  a  horizontal  beam,  the  top 
fibers  are  shortened  and  the  bottom  fibers  are  lengthened  by 
the  bending  stresses.     These  changes  in  length,  or  deforma- 
tions,  bend   the  beam  which  was  originally  straight  into  a 
curved   shape;   the   top   surface  becoming   concave,   and   the 
bottom  convex.     The  neutral  axis    of  the  beam  in  its  bent 
condition  is  called  the  elastic  curve.     The  vertical  distance 
through  which  a  point  on  the  elastic  curve  has  been  moved  by 
this   bending  is  called  the  deflection.     The  distortion  of  the 
fibers  by  the  shearing  stresses  also  contributes  to  the  deflection, 
but  the  additional  amount  is  generally  too  small  to  have  any 
practical  value.     The   structural    engineer   frequently   desires 
to  determine  the  deflection  of  a  beam  or  girder  at  one  or  more 
points  in  its  length.     This  in  itself  makes  a  study  of  deflections 
desirable;  but  a  more  important  use  for  the  theory  involved 
is  its  application  to  statically  indeterminate  structures. 

There  are  several  methods  by  which  deflections  caused  by 
bending  may  be  determined.  In  the  oldest  and  most  widely 
known  method,  the  second  differential  equation  of  the  elastic 
curve  is  derived.  This  equation  must  be  integrated  twice 
before  the  deflection  at  any  point  may  be  found.  The  method 
is  long  and  greatly  involved,  except  for  the  simplest  conditions 
of  loading.  The  simpler  and  less  known  method  of  area- 
moments  establishes  a  relation  between  a  tangent  to  the 
elastic  curve  and  the  bending  moment  diagram. 

137.  Method  of  Area -moments. — Let  A  and  B  (Fig.  127) 
represent  any  two  points  on  the  neutral  axis  of  a  beam,  which 
is  bent  by  any  arrangement  of  loading.     Through  A  and  B 
draw    the    tangents  AD  and  EC  intersecting  at  C,  and  the 

208 


SEC.  I 


DEFLECTION    OF   BEAMS 


209 


normals  AI  and  BI  intersecting  at  7.  Then  Z.4L8  =  /.BCD 
—  $.  Let  QPRS  represent  the  bending  moment  diagram  for 
the  portion  of  the  beam  between  A  and  B.  Let  EFHG  repre- 
sent an  element  of  the  beam  between  two  right  sections  EG  and 


FIG.  127. 


FIG.  128. 


FH  (drawn  to  a  larger  scale  in  Fig.  128)  which  were  parallel  and  a 
distance  ds  apart  before  the  element  was  bent  by  the  bending 
moment  M.  Let  r  (Fig.  128)  represent  the  radius  of  curvature 
of  the  neutral  axis  for  this  element.  The  fiber  at  the  neutral 
axis  remains  unchanged  in  length,  while  the  fiber  KL  at  a 


2IO  THEORY    OF   FRAMED    STRUCTURES  CHAP.  V 

distance  y  below  the  neutral  plane  has  been  increased  in  length 
from  ds  =  rd<f>  to  (r  +  y)d(f>.     Hence  the  total  strain  (change  in 

length)  in  the  length  ds  is  yd<f>  and  the  unit  strain  is  ~  —     Let  / 

ds 

represent  the  unit  stress  on  the  fiber  KL  ;  and  let  E  represent  the 
modulus  of  elasticity. 
' 


ds 

Let  7  =  the  moment  of  inertia  of  the  cross-section  about  the 
neutral  plane;  then 

My 
J  '      I 

Eydcf>       My 

whence  —  ^  —  =  —  ^- 

ds  I 

Mds 
d*=   -w 

BMds 


and 


CB,          C 
=    I    d^>  =    I 

JA  JA 


If  the  beam  in  its  natural  state  is  straight  (not  arched)  and 
is  properly  designed,  the  curvature  will  be  so  slight  that  ds  may 
be  replaced  by  dxt  allowing  the  integration  to  be  made  hori- 
zontally between  A  and  B  instead  of  along  the  path  of  the 
elastic  curve.  Then 

BMdx 

= 


If  the  beam  is  homogeneous  and  has  a  uniform  cross-section, 
E  and  7  are  constants,  and  the  equation  may  be  written  thus: 

i  r* 
*  =  Ei)AMdx  (I) 

The  expression  Mdx  represents  the  area  of  the  cross-hatched 
element  in  bending  moment  diagram.  Hence  the  integral 

CB 
expression  I    Mdx    is   the   area   of   the   M-diagram   between 

«/A 

the  ordinates  RS  and  PQ,  and  if  this  area  is  divided  by  El,  the 
quotient  is  the  angle  <£.  If  M  is  expressed  in  inch-pounds,  the 
area  Mdx  is  expressed  in  inch2-pounds.  If  E  is  expressed  in 
pound /inches2,  and  7  in  inches4;  then  El  is  also  expressed  in 


SEC.  I  DEFLECTION   OF  BEAMS  211 

inch2-pounds,  and  the  angle  0  is  a  ratio.  In  any  practical  beam 
<j>  is  comparatively  very  small;  hence,  when  the  tangent  CB 
(Fig.  127)  is  horizontal,  the  ratio  <£  may  be  taken  as  the  slope  of 
the  tangent  AD.  Likewise,  when  AD  is  horizontal,  the  ratio 
0  may  be  taken  as  the  slope  of  the  tangent  CB.  From  this 
analysis  the  first  principle  may  be  deduced. 

138.  First  Principle. — //  tangents  are  drawn  through  any  two 
points  on  the  elastic  curve  of  a  homogeneous  beam  of  uniform 
cross-section,  the  angle  which  one  tangent  makes  with  the  other 
tangent  equals  the  area  of  the  M.-diagram  between  the  two  points, 
divided  by  EL 

Now  imagine  that  the  unstrained  position  of  the  beam  was  in 
the  direction  AD,  and  that  the  beam  was  subsequently  bent  so 
that  the  point  D  moved  to  B\  the  point  A  remaining  station- 
ary. This  movement  is  caused  by  the  bending  of  all  the  ele- 
ments from  A  to  B.  The  bending  of  the  element  EFGH  causes 
the  point,  in  its  travel  from  D  to  B,  to  move  a  distance  dt  = 
xd(f>.  Since  the  curvature  is  comparatively  small,  the  path  of 
the  point  moving  from  D  to  B  deviates  but  slightly  from  the 
straight  line  DB.  Hence 

f*B  (*B  T      (*B 

t  =        dt=    I    xd4>  =  ^      Mxdx  (2) 

JA  JA  EIjA 

The  distance  DB  =  t  is  called  the  tangential  deviation;  since  it 
represents  the  distance  through  which  the  point  B  has  been 
displaced  by  the  curvature  of  the  beam,  when  AD  is  assumed  as 
the  original  position. 

In  Eqs.  (i)  and  (2),  /  is  the  gross  moment  of  inertia  of  the 
cross-section.  No  deductions  are  made  for  holes,  as  is  the  case 
when  the  strength  of  a  beam  is  being  computed. 

The  expression  Mxdx  represents  the  moment  of  the  elemental 
area  Mdx  about  the  ordinate  through  B.  Hence  the  integral 

Xjy 
1  Mxdx  represents  the  moment  of  the  area  QPRS 

about  the  ordinate  through  B,  and  is  called  the  area-moment  of 
QPRS  about  B.  The  area-moment  is  expressed  in  inches3- 
pounds,  when  M  is  expressed  in  inch-pounds.  Since  El  is 
expressed  in  inches 2-pounds,  the  tangential  deviation  t  is 
expressed  in  inches.  The  second  principle  may  now  be  stated. 


212 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  V 


139.  Second  Principle.  —  If  the  tangent  to  the  elastic  curve  is 
drawn  through  any  point  A,  the  tangential  deviation  at  any  other 
point  B  may  be  obtained  by  finding  the  area  of  the  M-diagram 
between  ordinates  through  A  and  B;  and  dividing  by  El  the 
moment  of  this  area  about  the  ordinate  through  B. 

Let  /  represent  the  centroid  of  the  area  QPRS,  and  let  k 
be  the  distance  from  /  to  the  ordinate  through  B.  Then 

area 


El 
Since,  from  the  first  principle 

area  QPRS 

~W'        "* 

then  /  =  k(f> 

Hence  the  tangents  to  the  elastic  curve  at  any  two  points  A 
and  B  intersect  on  the  ordinate  through  the  centroid  of  the  M- 
diagram  included  between  the  ordinates  through  A  and  B. 

SEC.  II.     SIMPLE  BEAMS  OF  UNIFORM  CROSS-SECTION 
140.  The  beam  in  Fig.  129^  is  a  2  by  i  in.  piece  of  wood  laid 


PIG.  i2pa. 

flatwise.  /  =  %  in.4  E  =  1,500,000  lb./m.2  Hence  El  = 
250,000  in.2-lb.  The  M-diagram  is  PQS.  The  deflection  A 
under  the  load  will  be  determined  in  several  ways,  by  drawing 


SEC.  II  DEFLECTION   OF  BEAMS  213 

the  tangent  to  the  elastic  curve  through  different  points  as 
shown  in  Figs.  i2ga-b-c.  Considerable  time  and  labor  may  be 
saved  by  exercising  good  judgment  in  choosing  the  most 
advantageous  point  in  the  elastic  curve  through  which  the 
tangent  is  to  be  drawn. 

In  Fig.  i2ga  the  tangent  to  the  elastic  curve  ATB  is  drawn 
through  T.  The  deflection  A  is  quickly  found  after  the  tan- 
gential deviations  t\  and  t%  have  been  computed. 


Where  Mi  is  the  bending  moment  at  any  distance  x  from  the 
ordinate,  on  which  the  tangential  deviation  is  required.     Hence 

Mi  =  45*. 


The  origin  for  /2  is  at  B,  hence  Mz  =  i$x,  and 

29,160 
~KT 


A  =  t\  -f-  —  (fe  —  h)  =  y''~~  =  — y?/     —  =  0.039  in. 
24  El         1,500,000 

The  expression   [  Mixdx  represents  the  area-moment  of  PQV 
about  P,  hence 

.  =  ^70X3X4)=^°.  .. 

Likewise,  the  expression  ^M^xdx  represents  the  area-moment 
of  SQV  about  5,  hence 

>o,i6o 


/  X 

==  EI^7°  X  9  X  12) 

Thus  it  is  clear  that,  when  the  M-diagram  can  be  conveniently 
divided  into  portions  whose  areas  and  centroids  are  easily 
found,  a  semigraphic  or  geometric  solution  can  be  quickly 
made.  The  area  of  the  M-diagram  to  be  considered  in  each 
case  is  included  between  two  ordinates.  One  ordinate  passes 
through  the  point  of  tangency,  on  the  other  ordinate  the  tan- 
gential deviation  is  found;  and  the  moment  of  this  area  is  taken 
about  the  latter  ordinate. 

In  Fig.  1296  the  tangent  is  drawn  through  A.     The  area- 


214 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.   V 


moment  for  /4  is  PQS  about  5;  and  for  /3,  the  area-moment  is 
PQV  about  QV. 


E/270  X  3(18  +  2)  =  16,200 


Z7<"°  X  3  X  ,) 


II,34O  —   I,62O 

A  =  -  -  = 

ILL 


,    f 

before. 


SEC.  II  DEFLECTION    OF   BEAMS  215 

In  the  algebraic  solution,  the  origin  for  /4  is  at  S.  M  =  i$x 
for  values  of  x  between  o  and  18,  and  M  =  1500  —  60  (x  —  18)  = 
i  ,080  —  45#,  for  values  of  x  between  18  and  24,  hence 


f 
lji 


29,160  +  16,200  _  45,360 
~~W~  El 

The  origin  for  /3  is  at  F,  hence  M  =  45  (6  —  x),  and 

,      _  1,620 


45  f  6 

^3   =    777  I      ^*X  ~   X  ' —  777- 

Eljo  El 

The  geometric  solution  is  considerably  shorter  when  M  is  not  a 
continuous  function  of  x  as  in  the  case  of  J4. 
In  Fig.  129^  the  tangent  is  drawn  through  B. 

^["270  X  3  X  4  =         3>24o]  =  32,400 
E/|_27o  X  9(6  +  6)  =  29,160]  El 


-  3  7      .   24,300 
—  -  h  —  —  ^j^- 

4  rLL 

24,300    —     14,580  0,720 

^  v? 


141.  Maximum  Deflection.  —  ^Let  X  (Fig.  129^)  represent  the 
point  of  maximum  deflection.  Since  the  tangent  through  X  is 
horizontal,  Amax  =  ti  =  /8-  Let  KL  be  the  ordinate  in  the  M- 
diagram  at  the  point  of  maximum  deflection,  and  let  LS  =  a. 
Then  KL  =  i$a.  Let  <£  represent  the  angle  which  the  tangent 
through  B  makes  with  the  horizontal  tangent  through  X,  then 


Z.BID  =  </> 
and  240  =  /5 


*  = 

24 
also  =  area 


E/ 

hence  7.5a2  =  1,350 

a  =  13.42 


2l6 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  V 


Since  the  centroid  of  the  triangular  area  KLS  is  on  the  ordinate 
through  /, 


ID  =  -a 

3 

ts  =  -a(f> 
3 


35°  \  _ 
El 


or 


-m 

[area-moment  0/1  _. 
KLS  about  S  J 


El 


_    I2,C 

~~EI~ 

5#3  _  12,078 
EJ  =    ~ET 


in. 


KLS  afow/ 

The  distance  a  might  also  be  found  by  equating  the  values  of  /7 
and  /8  without  any  reference  to  the  tangent  through  B. 

The  general  expression  will  be  developed  for  the  deflection 
of  .a  simple  beam  /  in.  long  when  supporting  a  single  concen- 


TT~ 


FIG.  129^. 


trated  load  of  P  Ib.  at  any  distance  kl  from  the  left  support, 
(Figs.  130  and  131).  The  deflection  A  is  found  at  T,  a  distance 
cl  from  the  left  support. 

When  c  <  k. 
In  Fig.  130  the  tangent  CD  is  drawn  through  T. 


equals  the  area-moment  of  KQSL  about  61  which  is  the 


SEC.  II 


DEFLECTION    OF  BEAMS 


217 


area-moment  of  PQS  about  5,  minus  the  area-moment  of 

PKL  about  S. 

B 


£ Ci  -- 

<-—kl ><- 


FIG.  131 


kl 


PI* 


—  (l    -   k)(2k   ~   k*   ~      C2  + 

JL     V        /  i  j  n  o         *  -I  o         i 

A  =  -^^(2ck  —  $ckz  —  c6  -\-  ck6  -\- 
When  c  >  k 


(3) 


2l8  THEORY    OF   FRAMED    STRUCTURES  CHAP.   V 

In  Fig.  131  the  tangent  CD  is  drawn  through  T  as  before. 

PP 
**  =  6EI 

PP 


A  =  /8  +  c(t*  -  ts) 

p/3 

A  =    Z(*ck  -  3c*k  -  ks  +  c*k  +  c&)  (4) 


The  deflection  at  the  load  may  be  obtained  from  either  Eq.  (3) 
or  (4).  Since  c  =  k  for  this  condition,  either  equation  reduces 
to 


Let  F  represent  the  expression  in  the  parenthesis  of  equation 
(3)  when  c  is  less  than  k,  and  the  expression  in  the  parenthesis 
of  Eq.  (4)  when  c  is  greater  than  k,  then  in  general 


The  values  of  F  for  various  values  of  c  and  k  are  given  in 
Table  I;  or  may  be  found  from  Fig.  132.  In  Eq.  (5)  /  is  the 
length  of  the  beam  in  inches,  and  P  is  the  load  in  pounds  at  any 
distance  kl  from  the  nearer  support.  A  is  the  deflection  in 
inches  at  any  distance  cl  from  the  same  support.  E  is  the 
modulus  of  elasticity  in  pounds  per  square  inch,  and  /  is  the 
moment  of  inertia  of  the  constant  cross-section  of  the  beam 
about  the  neutral  axis,  measured  in  inches4. 

142.  Point  of  Maximum  Deflection.  —  The  maximum  deflec- 
tion occurs  in  the  longer  segment  of  Fig.  131  where  c  is  greater 
than  k,  and  at  the  point  where  the  tangent  through  T  is  hori- 
zontal; hence  the  value  of  c  for  ATOaa;.  may  be  found  by  equating 
the  expressions  for  /3  and  /4,  whence 
c  =  i  -  i 


The  values  of  c  and  the  corresponding  values  of  F  for  Amaa!. 
are  also  given  in  Table  I. 

Since  the  limits  of  k  are  o  and  0.5,  all  values  of  c  for  maximum 
deflection  will  fall  between  0.4227  and  0.5.  Hence  the  point 
of  maximum  deflection  for  a  single  load  is  between  the  load 


SEC.  II 


DEFLECTION    OF   BEAMS 


219 


and  the  center  of  the  span,  and  always  relatively  near  the 
center.  The  most  eccentric  loading  which  a  simple  beam  of 
uniform  cross-section  and  span  /  can  experience,  occurs  when  a 


fa/aes  ofc 
FIG.  132. 


single  load  is  adjacent  to  one  of  the  supports,  and  k  is  on  the 
point  of  becoming  zero.  Under  this  condition  the  point  of 
maximum  deflection  cannot  be  at  a  distance  greater  than  0.07 73^ 
from  the  center  of  the  span.  Any  second  load  applied  to  the 


22O 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  V 


beam  must  necessarily  throw  the  point  of  maximum  deflection 
nearer  the  center.  Hence  the  point  of  maximum  deflection  of  a 
simple  beam  of  uniform  cross-section,  loaded  in  any  manner, 
will  be  near  the  center  and  not  more  than  0.0773  °f  its  length 
from  the  center. 

143.  A  2o-in.  65~lb.  I-beam  supports  two  loads  of  30,000  Ib. 
each  (Fig.  133).  Since  the  loads  are  symmetrically  placed, 
the  elastic  curve  and  M-diagram  are  symmetrical  about  the 
center.  The  tangent  to  the  elastic  curve  at  the  center,  drawn 

FT 


through  T,  is  horizontal  and  A  =  /.     E  =  29,000,000  lb./in2., 
I  =  1,169.5  in4.;  hence  El  =  33,915,500,000  in2.-lb. 

Area-moment  of  PQSU  about  P. 
areaPQF         150,000  X  2.5  X  3.33  =  1,250,000 
area  QSUV     150,000  X  7-5  X  8.75  =  9,843,750 

11,093,750  ft3.-lb. 
A  =  /  =  "»Q93,75o  X  1,728  =       6    in 


When  the  length  of  a  beam  is  expressed  in  feet,  and  the  loads 
are  expressed  in  pounds,  the  area-moment  will  be  expressed  in 
foot3-pounds;  and  the  factor  1,728  is  introduced  if  El  is  ex- 
pressed in  inch2-pounds. 

El  may  be  expressed  in  foot2-pounds  by  dividing  by  144, 
whence 
El  =  235,521  ft.2-lb.,  then 

=     =  11,093,750  ft  3-lb.  =  ft  in 

235,521  ft.2-lb. 

The  deflection  at  the  center  may  be  found  from  Table  I. 


SEC.  II 


DEFLECTION    OF   BEAMS 


221 


k  =  0.2  and  c  =  0.5;  therefore  F  =  0.071  for  each  load,  then 

2PP  t  60,000(25  X  I2)30.07I 

A  =  — -  X  0.071  =  -  -    =  0.565  in. 

6EI  6  X  33,915,500,000 

144.  Deflection  Under  Uniform  Load. — The  beam  in  Fig. 
134  supports  a  uniform  load  and  the  M-diagram  PQS  is  a 


PIG.  135. 

parabola.  The  maximum  deflection  is  at  the  center  of  the 
span.  The  tangent  to  the  elastic  curve  at  C  is  horizontal,  and 
A  =  /.  Elt  =  the  area-moment  of  SQV  about  S.  The  area 
SQV  is  two- thirds  the  area  of  the  rectangle  QPSV,  and  the 
centroid  of  the  area  SQV  is  five-eighths  of  VS,  hence 

A  -  I  -  ^(I9,2oo  X  8  X  |  X  5  X  1.7,8)  =  "*ffi°°° 


222 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  V 


o 

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00 

M 

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6  6 

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M 

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00      tv. 

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00       M 

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SEC.  II 


DEFLECTION    OF   BEAMS 


223 


In  Fig.  135  the  span  is  /  in.,  and  the  uniform  load  is  w  Ib.  per 
inch.     The  deflection  at  any  distance  cl  from  A  will  be  found. 

At  any  distance  x  from  either  support,  M  =  —  x  (I  —  x) . 


if 


-  I    Mxdx  =  — (4c3 
24 

f(l-c)l  ^4 

- 1  Mxdx  =  — (i  — 
rjo  24 

A  = 


Let  W  =  the  total  uniform  load,  then  W  =  wl,  whence 

A  =  ~=J(c  -  2C3  -f-  c4)  (6) 


Let  /  represent  the  expression  in  the  parenthesis  of  Eq.  (6) 
then 

WP 


A  = 


(7) 


The  values  of  /  for  various  values  of  c  are  given  in  Table  II;  in 
which  /  is  the  length  of  the  beam  in  inches,  W  is  the  total  uni- 
formly distributed  load  in  pounds  and  A  is  the  deflection  in 
inches  at  any  distance  cl  from  the  nearer  support.  E  is  the 
modulus  of  elasticity  in  pounds  per  square  inch,  and  /  is  the 
moment  of  inertia  of  the  constant  cross-section  of  the  beam 
about  the  neutral  axis,  expressed  in  inches.4 

For  Amaa;.,  c  =  0.5,  hence 

WP  x 


TABLE  II 


c 

/ 

c 

J 

0.05 

o  .  0498 

0.30 

0.2541 

O.  IO 

0.0981 

o-35 

0.2793 

0.15 

0.1438 

0.40 

o.  2976 

0.20 

0.1856 

0-45 

0.3088 

0.25 

o.  2227 

0.50 

0.3125 

224 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  V 


145'  Deflection  for  Load  of  Uniformly  Varying  Intensity. — 

The  beam  in  Fig.  136  supports  a  load  of  uniformly  varying 
intensity.  The  total  load  is  W  lb.,  and  the  length  of  the  beam 
is  /  in.  The  bending  moment  at  any  distance  x  from  A  is 

Ml  =  ^(I2x  -  x3) 
The  bending  moment  at  any  distance  x  from  B  is 


+ d >L< (hcjl 


FIG.  136. 


w 


EI 


=     ^    f 

EIJ, 


Wl3 
iSoEI 


(7   —  30C2  +  20C3  +  I5C4  —  I2C5) 


A  =  /i  +  c(h  -  /i)  = 


Wl* 


-  ioc 


iSoEI 

The  value  of  c  for  Awax.  may  be  found  by  equating  t\  and  /2, 
whence 

i5c4  -  3oc2  =  -7 
c  =  0.519 

=     Wl3  Wl3 

'  iSoEI  ~  o.oi3i£7 

The  intensity  of  the  load  in  Fig.   137  increases  uniformly 
from  each  support  to  the  center  of  the  span.     The  total  load  is 


SEC.  II 


DEFLECTION    OF  BEAMS 


225 


W  lb.,  and  the  length  of  the  beam  is  /  in.     The  bending  moment 
at  any  distance  x  between  the  end  and  center  is 

jp- 

1  ~~  oT2  3 

The  bending  moment  at  any  distance  x,  when  x  is  greater 
than  Vo/  is 


<---« ->k (l-c)l 


W 

~ 

T 


FIG.  137. 


i  r*  i  r( 

=        I   Jf  lO^t;  +  —  I 


A  = 


W/3 


T 

4001  tL 


is 


226 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  V 


in  which  c  may  have  any  value  between  o  and  J^.     The  value 
of  c  for  Amaa;.  may  be  found  by  equating  t\  and  /2  whence, 


=  m* 

~  6oEI 

146.  Deflection  for  Several  Concentrated  Loads. — A  24-in. 
by  8o-lb.  I-beam  supports  three  loads  (Fig.  138).     The  linear 


10000^ 


5000' 


/COO1 


10800 


-  -10'-- 


-  -  -  15'-  ----><--  -/<?'-->  <-  -  -  -/5-  -  -  -> 


5^.00 


A  max. 


dimensions  of  the  beam  are  expressed  in  feet  and  the  units  in 
which  E  and  7  are  usually  expressed  will  be  changed  accord- 
ingly. 7  =  2,087  m-4  E  ~  29,000,000  lb./in.2  Hence  El  = 
60,532,000,000  in.2-lb.  or  420,400,000  ft.2-lb.  The  tangent  is 
drawn  through  C  at  the  center  of  the  span.  The  deflection 
due  to  the  weight  of  the  beam  will  be  considered  later. 

Area-moment  about  P. 

area  PQY  108,000  X  5  X  6.67  =  3,600,000 
area  YQW  108,000  X  7.5  X  15  =  12,150,000 
area  QRW  120,000  X  7.5  X  20  =  18,000,000 


33,750,000 
420,400,000 


33,750,000  ft.3-lb. 

=  0.0803  ft- 


SEC.  II  DEFLECTION    OF   BEAMS  227 

Area-moment  about  U. 

area  USV  78,000  X  7.5  X  10  =  5,850,000 
area  SVW  78,000  X  7.5  X  18.33  =  7,150,000 
area  SR W  120,000  X  5  X  21.67  =  13,000,000 

26,000,000  ft.3-lb. 

26,000,000  ,  0  f  . 

h  =  -  -  =  0.0618  ft. 

420,400,000 

A  =  — — 2  =  0.071  ft.  =  0.85  in. 

2 

The  maximum  deflection  caused  by  the  three  loads  is  at  T, 
where  the  tangent  is  horizontal,  and  the  ordinate  in  the  M- 
diagram  is  KL.  Let  LW  =  a  and  let  <f>  be  the  angle  made  by 
the  two  tangents;  then  0  represents  the  slope  of  the  tangent 
through  C.  The  beam  is  50  ft.  long 

i  /i  —  /2       0.0185 

hence  <f>  =  -        -  =  »  =  0.00037 

50  5° 

also  0  =  areagWL 

hence  area  KRWL  =  <f>  El  =  155,000  ft.2-lb. 
whence  a  =  1.3  ft. 

The  centroid  of  the  area  KRWL  is  approximately  24.35  ft- 
from  P,  and  the  area-moment  of  KRWL  about  P  is 

155,000  X  24.35  =  3,774,ooo  ft.3-lb. 
A          -  /       3,774,ooo  _  29,976,000 

<*•*•*•   —  rl         =r= —   — 


A....  =  ,_  =  ft   =  Q  86  .n 

420,400,000 

Although  the  loads  are  eccentric,  it  is  clear  that  there  is  practi- 
cally no  difference  between  the  deflection  at  the  center  and  the 
maximum  deflection. 

The  deflection  at  the  center  may  be  found  from  Eq.  5,  page 
218.     The  coefficients  F  are  given  in  Table  I.     c  =  0.5;  k  = 


228   , 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  V 


0.2  for  the  load  at  A]  0.5,  for  the  load  at  B\  and  0.3  for  the 
load  at  C;  hence 

10,000  X  so3  . 

— - — *--  X      0.071  =    14,792,000 

5,000  X  5Q3 
6  " 


X      0.125  =    13,021,000 


X 


_ 


_       2,062,000 


29,875,000  ft3.-lb. 


A  = 


=  0.071  ft.  =  0.85  in. 


*7  5000* 


- 30 '- »U- 15 '-  -  - 


PIG.  139. 

The  deflection  at  the  center,  due  to  the  weight  of  the  beam, 
may  be  found  from  equation  7,  page  223.  The  coefficient  of 
/  when  c  =  0.5  is  0.3125.  W  =  50  X  80  =  4,000. 


A  =       4,ooo  X5Q3 


24  X  420,400,000 


= 


SEC.  II  DEFLECTION    OF  BEAMS  229 

The  total  deflection  at  the  center  is  0.85  +  0.19  =  1.04  in., 
which  in  this  case  may  be  assumed  without  appreciable  error 
as  the  maximum  deflection.  A  deflection  of  ^60  of  the  span 
is  considered  not  excessive. 

147.  Deflection  for  Uniform  and  Concentrated  Loads.— 
In  Fig.  139,  the  tangent  is  drawn  through  C  at  the  center  of  the 
span.  Assume  that  El  is  expressed  in  foot2-pounds.  The 
M-diagram  under  the  uniform  load  cannot  be  accurately 
divided  into  triangles,  and  an  integration  is  necessary  if  an 
accurate  solution  is  desired.  A  sufficiently  accurate  solution 
for  all  practical  purposes  may  be  obtained  by  the  geometric 
process  by  dividing  the  area  QBDFJ  by  vertical  ordinates  into 
strips,  so  narrow  that  their  areas  may  be  considered  trapezoidal. 
The  accurate  method  by  integration  is  given  below.  Let  MI 
represent  the  bending  moment  under  the  uniform  load  at  any 
distance  x  from  the  left  support;  and  If 2,  the  bending  moment 
under  the  uniform  load  at  any  distance  x  from  the  right 
support;  then 

Mi  =  —loox2  +  5,0000;  —  2,500 
and  Mi  =  —  loox2  +  5,8000;  —  24,100 

Area-moment  about  A 
area  ABQ        20,000  X  2.5  X  3.33  =        166,667 

/»2? 

area  BENQ   \     M^xdx  =   18,446,266 

18,612,933  ft3.-lb. 
ti  =  18,612,933  ft 

Area-moment  about  H 

area  GHI         20,000X2      X    2.67=  106,667 

area  G/J          20,000  X  7.5  X    9       =  1,350,000 

area  FGJ         50,000  X  7.5  X  14       =  5,250,000 

aresiFENJ    I     M 2  x  dx  =  10,332,600 

17,039,267  ft.Mb. 
17,039,267  f 

~ET 
The  deflection  at  the  center  is 

/i  +  fe       17,826,100  - 

~2~  ~lH~ 


230  THEORY    OF   FRAMED    STRUCTURES  CHAP.   V 

The  slope  of  the  tangent  is 

_  ti  -  /2  _  29,142 
~5T        ~ET 

Let  KL  represent  the  ordinate  in  the  M-diagram  at  the  point  of 
maximum  deflection,  then 

area  KENL 
0=         ~ET 

therefore      area  KENL  =  29,142 
whence  LN  =  0.488  ft. 

The  area-moment  of  KENL  about  A  is 

29,142  X  26.756  =  779,723 

A  -      /  779,723       __       I7,833,2H 

El  EI~ 

The  area-moment  of  KENL  about  H  is 

29,142  X  27.244  =  793,945 
A          _  ,     ,    793,945  __  17,833,212 
El  El 

When  the  tangent  is  drawn  to  the  elastic  curve  at  the  right 
end  of  the  uniform  load,  the  tangential  deviations  fa  and  /4,  at 
the  left  and  right  supports  respectively,  may  be  determined  by 
the  geometric  process;  for  if  a  straight  line  be  drawn  from  B  to 
F,  the  area  of  the  M-diagram  BDF  has  all  the  properties  of  the 
M-diagram  B'D'F'  for  a  beam  30  ft.  long  when  uniformly 
loaded  with  200  Ib.  per  foot  (Fig.  1396).  See  Article  60. 

Area-moment  about  A 

area  ABQ        20,000  X    2.5  X    3.33       =        166,667 
area  BQJ         20,000  X  15     X  15  =    4,500,000 

area  FBJ         50,000  X  15     X  25  =  18,750,000 

area  BDF        22,500  X  30     X  %  X  20  =    9,000,000 

32,416,667  ft3.-lb. 

Area-moment  about  H 

area  GHI  20,000  X  2  X  2.67  =  106,667 
areaG/J  20,000  X  7.5  X  9  =  1,350,000 
area  FGJ  50,000  X  7.5  X  14  =  5,250,000 

6,706,667  ft3.-lb. 


SEC.  Ill  DEFLECTION    OF  BEAMS  231 

The  slope  of  the  tangent  is 

/s  —  /4  _  476,315  _  area  KFJL 
~~S4~       ~Ef"  ~EI~ 

The  area  EFJN,  when  considered  as  four  trapezoidal  areas, 
each  2  ft.  wide,  is  446,440;  hence  the  approximate  area  of 
KENL  is 

476,3J5  -  446,400  =  29,915 

from  which  we  find  that  the  ordinate  KL  is  located  about  0.5 
ft.  to  the  left  of  the  center  of  the  beam  as  before. 


SEC.  III.     MAXWELL'S  THEOREM  OF  RECIPROCAL 
DISPLACEMENTS 

148.  Maxwell's  Theorem  of  Reciprocal  Displacements  estab- 
lishes a  mutual  relation  between  any  two  points  in  a  structure. 
This  theorem,  when  considered  in  connection  with  the  deflec- 
tion of  beams,  may  be  stated  as  follows:  If  the  load  P  at  A 


FIG.  140.  FIG.  141.  FIG.  142. 

(Fig.  140)  causes  a  deflection  A2  at  B,  and  the  load  P  at  B 
(Fig.  141)  causes  a  deflection  A3  at  A;  then,  according  to 
Maxwell's  theorem,  A2  =  A3.  Let  Figs.  140,  141  and  142 
represent  the  deflections  of  a  beam  when  the  loads  are  applied 
gradually.  When  A  (Fig.  140)  has  received  its  full  load,  the 
work  done  is  J^PAi.  With  a  full  load  P  at  A,  let  another  load 
P  be  gradually  added  at  B.  The  deflections  as  shown  in  Fig. 
142  will  result.  The  point  A,  with  the  full  load  P,  moves 
through  the  additional  distance  A3;  and  the  point  B  moves 
through  the  additional  distance  A4,  as  the  load  P  is  gradually 
applied  at  B.  Hence  the  total  work  done  is 

-PAi+PA3  +  -PA4 

2  2 

If  B  is  loaded  first  and  then  A  is  loaded,  the  total  work  done  is 
-PA4+PA2  +-PAi 

2  2 


232 


THEORY  OF  FRAMED  STRUCTURES 


CHAP.  V 


The  total  amount  of  work  done  in  each  case  is  the  same,  hence 

A2  =  A3 

Maxwell's  law  may  be  verified  by  Table  I.  When  A  and  B  are 
on  the  same  side  of  the  center,  the  values  of  k  and  c  for  Fig. 
140  become  interchanged  for  Fig.  141.  For  example,  F  = 
0.0658  when  k  =  0.2  and  c  =  0.3;  likewise,  F  =  0.0658  when 
k  =  0.3  and  c  =  0.2.  When  A  and  B  are  on  opposite  sides  of 
the  center  the  application  is  made  as  follows:  In  Fig.  140  let 
k  =  0.3  and  c  =  0.8;  then  in  Fig.  141,  k  =  0.2  and  c  =  0.7; 
whence  F  =  0.0522  in  each  case.  Maxwell's  law  renders 
excellent  service  in  the  solution  of  statically  indeterminate 
structures. 

SEC.  IV.     CANTILEVER  BEAMS 

149.  The  beam  in  Fig.  143  supports  a  single  load  P  at  the 
free  end.     It  is  fixed  in  the  wall  at  A  in  such  a  way  that  the 


-PI 


FIG.  143. 

tangent   to  the  elastic  curve  at  A  remains  horizontal,  hence 
the  deflection  at  the  free  end  is 

P/3 


A  =  I  = 


&I 


The  negative  sign  indicates  that  the  elastic  curve  deviates 
below  the  tangent. 

The  beam  in  Fig.  144  supports  the  load  W  uniformly  dis- 
tributed.    The  M-diagram  is  SQV.     The  curve  SV  is  a  para- 


SEC.  IV  DEFLECTION    OF  BEAMS  233 

bola  with  the  vertex  at  5,  hence  the  deflection  at  the  free  end  is 


A  7-in.,  i5-lb.  I-beam  (Fig.  145)  supports  a  load  of  2,000  Ib. 


FIG.  145. 

El  =  1,050,000,000  in.2-lb.     The  tangent  is  drawn  through  C. 

—  12,000  X   1-5   X   2 

tl   =   - 


=  _Q 


1,050,000,000 

=  -12,000X3X4X1,728  _  _Q          in 


I  ,O5O,OOO,OOO 

A  =  2ti  +  h  =  —0.3552  in. 


234 


THEORY  OF  FRAMED  STRUCTURES 


CHAP.  V 


The  deflection  at  B  may  also  be  found  by  drawing  the  tangent 
through  either  A  or  B. 

A   cantilever  beam  is  shown  in  Fig.    146.     In  finding    t\t 


?000' 


10000$ 


5000 


;----  /z '----. 


positive  and  negative  areas  are  encountered  in  the  M-diagram. 
These  may  be  treated  in  one  of  two  ways.  The  point  of  zero 
bending  moment  at  I  may  be  determined,  and  the  areas  PIW 
and  IQV  treated  separately;  or  the  area  WIV  may  be  included 
with  both  positive  and  negative  areas  as  follows : 


SEC.  V  DEFLECTION    OF   BEAMS  235 

Area-moment  about  P. 

area          PVW -  30,000  X  3  X     2  =  -180,000 
area          WQV+  36,000  X  3  X    4  =  +432,000 
area          QVU  +  36,000  X  6  X  10  =  +2,160,000 
area         QRU  +  48,000  X  6  X  14  =  +4,032,000 

+  6,444,000  ft.3-lb. 
If  El  is  expressed  in  foot2-pounds  then 

_  6,444,000  , 
fc-       -^-ft. 

Area-moment  about  S 
area  ,SW       48,000  X  6  X  8  =  2,304,000  ft.3-lb. 


<6  =  — *  = 

30  EI 

Let  KL  represent  the  ordinate  in  the  M-diagram  at  the  point 
of  maximum  deflection,  then 
area  KRUL  =  138,000 
a    =  2.967  ft. 
KL  =  45>°33 

The  maximum  deflection  may  now  be  found  as  in  previous 
cases. 

SEC.  V.    BEAMS  WITH  VARYING  CROSS-SECTION 

150.  General  Expressions. — The  moment  of  inertia  of  beams 
having  uniform  cross-section  is  constant,  and  for  this  reason  / 
appears  outside  the  integral  sign  in  Eqs.  (i)  and  (2).  When 
the  cross-section  is  not  uniform  the  moment  of  inertia  varies, 
and  Eqs.  (i)  and  (2)  become 

* = i  (B—  (8) 

•  1~1     I  T  \        / 

EjA        I 

tBMxdx 


I 

In  order  to  perform  the  integration,  /  as  well  as  M  must  be 
expressed  as  a  function  of  x.  This  is  relatively  a  simple  matter 
when  the  beam  has  a  rectangular  cross-section  varying  uni- 
formly in  breadth  or  depth;  but  this  method  often  results  in 


236 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  V 


long  and  cumbersome  expressions  when  applied  to  structural 
steel  sections.  In  all  such  instances  the  geometric  method  is 
preferable. 

151.  Beams  with  Varying  Depth. — The  beam  in  Fig.  1470  is 
a  plate  girder.  The  %  in.  web  plate  is  24  in.  wide  at  the  ends, 
and  the  width  increases  uniformly  to  36  in.  at  the  center. 
Each  flange  is  composed  of  two  angles  5  by  3  J^  by  %  with  the 


3-in.  leg  against  the  web.  The  distance  back  to  back  of  angles 
is  the  width  of  the  web  plus  %  in.  Ordinates  in  the  M-diagram 
(Fig.  1476)  are  given  in  inch-pounds  every  3  ft.  An  I-diagram 
is  shown  in  Fig.  147^.  The  ordinates  represent  the  moment 
of  inertia  in  inches4  at  3-ft.  intervals.  Each  ordinate  in  the 
M-diagram  has  been  divided  by  the  corresponding  ordinate 

M 
in  the  I-diagram,  and  the  quotient  recorded  in  the  -^-diagram 

(Fig.  i47</).  The  ordinates  in  this  diagram  are  expressed  in 
pounds/inches.3  Since  the  girder  and  the  loads  are  symme- 

M 

trical,  the  -r  -diagram  is  symmetrical  about  the  vertical  ordinate 


SEC.  V  DEFLECTION    OF   BEAMS  237 

through  the  center  of  the  span;  the  maximum  deflection  is  at 
the  center,  and  the  tangent  to  the  elastic  curve  (not  drawn) 
at  the  center  is  horizontal;  hence  the  tangential  deviation  /  at 
the  left  support  equals  the  area-moment  A  BCD  about  A 
divided  by  E. 

Area-moment  about  A 

443.6  X  36  X  36  =  575.0°° 
748.5  X  36  X  72  =  1,940,000 
957.8  X  36  X  108  =  3,724,000 
917.0  X  36  X  144  =  4,754,0°° 
876.5  X  36  X  1  80  =  5,680,000 
837.2  X  18  X  204  =  3,074,000 

19,747,000  Ib./in. 
Win.   .    oAo- 

—    O.OO  in. 


ma>  .     9 

29,000,000  lb./m2. 

M 
When  the  ordinates  in  the  ^'diagram  are  computed  only 

for  the  ordinates  at  B  and  C,  and  AB  and  BC  considered  as 

straight  lines,  the  computations  are  as  follows: 
Area-moment  about  A 

957.8  X  54  X  72  =  3,724,000 
957-8  X  54  X  144  =  7,448,ooo 
837.2  X  54  X  1  80  =  8,138,000 

9,310,000 

19,310,000          ,    . 
Awa*.  =  -%&-        -  =  0.67  in. 
29,000,000 

M 

Thus  it  is  clear  that,  if  the  ordinates  in  the  -=-  -diagram  were 

relatively  close  together,  say  at  every  foot  or  closer;  or  even  if 
/  were  expressed  as  an  exact  function  of  x  in  Eq.  (9)  and  the 
integration  performed,  the  results  in  either  case  would  not 
differ  materially  from  those  obtained  above. 

152.  Beams  with  Cover  Plates.  —  The  plate  girder  in  Fig. 
148^  consists  of  a  24  by  %  web  plate,  four  angles  5  by  3^  by 
%  and  two  cover  plates  12  by  %  by  24  ft.  symmetrical 
about  the  center  line.  The  M-diagram  is  shown  in  Fig.  1486, 

M 

the  I-diagram  in  Fig.  148^,  and  the  -y-  -diagram  in  Fig. 


THEORY    OF    FRAMED    STRUCTURES 

Area  moment  about  A 

1.065.1  X  36  X    48  =  1,840,500 
631.6  X  18  X    84  =  955>°°o 
947.4  X  18  X    96  =  1,637,100 
947.4  X  54  X  144  =  7,367,000 

1.263.2  X  54  X  1 80  =  12,278,300 

24,077,900 


CHAP.   V 


?0000*  20000* 


20000" 


The  deflection  at  the  center  is 

A  -  24'°77'9°°  =  0.83  in. 

29,000,000 

The  solution  by  integration  may  be  obtained  as  follows:  Let 
M i  represent  the  bending  moment  for  values  of  x  between  o 
and  9;  and  M%  the  bending  moment  for  values  of  x  between  9 
and  1 8.'  Then 

Mi  =  30,0000: 

Mz  =  io,ooo#  +  180,000 
Then  for  values  of  x  between  o  and  6,  EI\  =  408,417,000  ft2.- 


SEC.  V  DEFLECTION    OF    BEAMS  239 

lb.  and  for  values  of  x  between  6  and  18,  EI2  =  688,750,000 
ftVlb.     The  deflection  at  the  center  expressed  in  feet  is 


A  =  /  =  ~-      M&dx  +  -r      M.xdx  +  -r       Mixdx 

&1\J    0  •C'-LzJ   6  tillj    9 

The  solution  by  integration  is  much  more  simple  in  this 
problem  than  in  the  preceding  one,  for  in  this  problem  /  is 
constant  between  certain  limits  of  x  and  is  therefore  not  a  func- 
tion of  x. 

The  geometric  treatment  by  area-moments  is  by  far  the 
simplest  and  best  method  now  known  for  obtaining  a  solution 
of  any  practical  problem  involving  the  deflection  of  beams. 


CHAPTER  VI 
RESTRAINED  AND  CONTINUOUS  BEAMS 

SEC.  I.     RESTRAINED  OR  FIXED  BEAMS 

153.  General  Considerations. — The   beam   in   Fig.  149   is 
considered  fixed  or  restrained  at  A ,  if  built  into  the  wall  in  such 


U 


FIG.  149. 


a  way  that  any  attempt  of  the  external  forces  to  rotate  the 
beam  at  A  is  successfully  resisted,  and  the  neutral  plane  of  the 
beam  in  its  original  position  AB  remains  tangent  to  the  elastic 
curve  at  A  when  the  beam  is  bent.  Under  these  conditions  the 
reactions  RI  and  Ri,  and  the  resisting  moment  Mi  at  A  present 

240 


SEC.  I  RESTRAINED   AND    CONTINUOUS   BEAMS  241 

more  unknown  quantities  than  can  be  determined  by  the 
principles  of  statics;  and  the  beam  is  statically  indeterminate. 
The  principles  of  deflections  render  their  most  helpful  service 
in  the  solution  of  problems  of  this  character.  The  ease  with 
which  problems  of  this  kind  are  solved,  depends  to  some  extent 
upon  the  manner  in  which  the  M-diagram  is  drawn;  for  it 
may  be  represented  in  three  ways  as  shown  in  Figs.  1496, 
I49C  or  149 d\  and  each  case  will  be  considered  separately. 

154.  Restraint  at  One  End — Concentrated  Load. — In  Fig. 
1496,  let  MI  and  M^  represent  the  bending  moments  at  A  and  C 
respectively.  The  bending  moment  at  B  is  zero.  The  line 
AB  is  tangent  to  the  elastic  curve  at  A,  and  the  tangential 
deviation  t  at  B  is  zero,  therefore 

i    CB 
t  =  —  I  Mxdx  =  o 


or  CB 

Mxdx  =  o 


.... 

Hence  the  area-moment  of  QVUTP  about  Q  is  zero,  or 

^2(3  X  4)  +  Mz(6  X  10)  +  Mi(6  X  14)  =  o 

yAf  i  +  6M  2  =  o 

From  statics        MI  =  —(540  X  12)  +  18^2 
and  Mi  =  6R2 

whence  —  45,360  -+-  i267?2  +  36^2  =  o 

Rz  =  280 
Ri  =  260 
Mi  =  —1,440 
and  If  2  =  i,  680 

In  Fig.  149*;,  the  M-diagram  is  drawn  in  parts;  QST  is  the 
M-diagram  for  the  reaction  at  B,  and  TPU  is  the  M-diagram 
for  the  load  at  C.  The  area  QST  is  positive  and  the  area  TPU 
is  negative.  The  area-moment  of  the  total  diagram  about  Q  is 
zero;  therefore 

18^2(9  X  12)  -  6,480(6  X  14)  =  o 

Rz  =  280 
TS  =  iSRz  =  5,040 

PV  =\X  5,040=  i  ,680 

o 

SU  =  5,040  —  6,480  =  —1,440 

16 


242  THEORY    OF    FRAMED    STRUCTURES.  CHAP.  VI 

In  the  two  preceding  solutions  no  speculation  was  made  as  to 
the  general  form  of  the  elastic  curve.  The  curve  ACB  might 
have  had  any  shape  whatsoever,  so  long  as  its  tangent  at  A 
passes  through  B.  It  is  not  always  wise  to  presume  upon  the 
general  form  of  the  elastic  curve  before  computations  are  made; 
but  in  the  present  simple  case  it  is  quite  safe  to  assume  that 
the  curve  is  concave  on  the  under  side  near  A ,  and  concave  ofi 
the  upper  side  at  C,  with  a  point  of  contraflexure  between. 
Hence  the  bending  moment  is  negative  at  At  positive  at  C 
and  zero  at  an  intermediate  point  7;  consequently  the  M- 
diagram  may  be  sketched  as  in  Fig.  149^.  In  finding  the 
area-moment,  the  area  TIV,  which  is  not  a  part  of  the  diagram, 
can  be  included  as  positive  area  with  QVI,  and  as  negative 
area  with  TIV. 

^2(3  X  4)  +  M*(6  X  10)  -  Mi(6  X  14)  =  o 

yMi  —  6M%  =  o 

From  statics  —M\=  —(540  X  12)  -f  i8R2 

and  M-2  =  6Rz 

whence  R%.=  280 

#1  =  260 

—  Mi  =  —1,440 

Mi  =  i, 680 

When  an  unknown  ordinate  in  the  M-diagram  is  represented 
by  a  symbol,  it  is  generally  better  to  assume  that  the  ordinate 
is  positive  as  in  Fig.  149^;  then  if  the  solution  shows  that  the 
ordinate  is  negative,  the  M-diagram  may  be  re-drawn  if 
desirable.  Frequently  the  M-diagram  may  be  constructed  to 
advantage,  as  shown  in  Fig.  1490. 

Only  two  independent  static  equations  can  be  written  for  the 
solution  of  a  system  of  parallel  forces.  In  the  present  problem 
there  were  three  unknown  quantities  to  be  determined,  hence 
one  elastic  equation  was  necessary  for  a  solution. 

155.  Restraint  at  One  End — Uniform  Load. — The  beam  in 
Fig.  1500,  fixed  at  A  and  simply  supported  at  B,  carries  a  total 
load  Wj  uniformly  distributed.  In  Fig.  1506  QTU  is  the  M- 
diagram  for  the  uniform  load  and  QTS  is  the  M-diagram  for  the 
reaction  at  B.  If  the  tangent  to  the  elastic  curve  is  drawn 


SEC.  I 


RESTRAINED   AND    CONTINUOUS  BEAMS 


243 


through  A,  the  tangential  deviation  t  at  B  is  zero,  hence  the 
area-moment  of  QSU  about  Q  is  zero. 


rs  -  -  in 

o 

The  M-diagram  may  now  be  drawn  to  scale  as  in  Fig. 


FIG.  150. 

the  resisting  moment  at  A  is 

SU  =  TS  -  TU  =  §  Wl  -  -  Wl  =  -  \  Wl 

02  O 

156.  Restraint  at  Both  Ends— Concentrated  Load.— The 

beam  in  Fig.  1510  is  fixed  at  each  end.  The  M-diagram  is 
sketched  in  Fig.  1516  by  assuming  all  ordinates  positive.  It 
cannot  be  drawn  to  scale  until  Mi,  Mi  and  M 3  are  known. 
There  are  four  unknown  quantities  involved  in  the  external 


244 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VI 


forces  acting  at  the  points  of  support,  R\,  R%,  Mi  and  Mz  for 
which  four  independent  equations  are  necessary.  The  two 
static  equations  may  be  written  thus, 

Rl  +  Rz  =  P  (i) 

Ms  =  klRi  +  Ml  =  (i  -  k)lRz  +  Mz  (2) 

Two  elastic  equations  are   necessary.     Let  <£  be  the  angle 


which  the  tangent  through  A  makes  with  the  tangent  through 
B\  then,  since  0  =  o,  the  area  of  the  M-diagram  between  A 
and  B  is  zero;  therefore 

2  2  3) 

The  line  AB  is  tangent  to  the  elastic  curve  at  B,  and  the 


SEC.  I  RESTRAINED   AND    CONTINUOUS  BEAMS  245 

tangential  deviation  at  A  is  t\  =  o;  therefore  the  area-moment 
of  the  M-diagram  about  TV  is  zero,  hence 


(4) 


A  third  elastic  equation  may  be  written  by  equating  to  zero 
the  area-moment  of  the  M-diagram  about  QP\  but  obviously 
this  equation  would  not  be  independent  of  Eqs.  (3)  and  (4). 
Eqs.  (3)  and  (4)  may  be  reduced  to 

kMi  +  (i  -  k)M*  +  Ms  =  o         (3a) 
k*Mi  +  (2  -  k  -  k*)Mz  +  (i  +  k)Mz  =  o        (4a) 
Solving  Eqs.  (i),  (2),  (30)  and  (40) 


Afi  =  -k(i  -  k)2Pl 
M2  =  -  k2(i-  k)Pl 

MS  =    2/^2(l    -   k)2Pl 

Since  the  limits  of  k  are  o  and  i,  it  is  clear  that  Mi  and  Mz 
are  negative  bending  moments  and  M$  is  a  positive  bending 
moment.  The  M-diagram  in  Fig.  151^  is  the  same  as  in  Fig. 
1516,  except  that  TQ  has  been  drawn  in  the  horizontal  position. 
Let 

SV  =  M3  =  M*  +  M& 
From  geometry 

M5  =  M2  +  (i  -  k)(Mi  -  Mz) 
hence  M4  =  Mz  -  M5  =  k(i  -  k)Pl 

If  the  beam  were  simply  supported,  not  fixed,  at  A  and  B,  the 
bending  moment  at  C  would  be  M*  =  k(i  —  k)Pl,  therefore 
TSQ  is  the  M-diagram  when  the  beam  is  not  restrained  at  the 
ends  by  MI  and  M*.  In  a  numerical  problem  the  M-diagram 
should  be  sketched  as  in  Fig.  1516,  and  Mi  computed  as  for  a 
simple  beam.  After  the  negative  moments  Mi  and  Mz  have 
been  determined,  the  trapezoid  TQPU  may  be  revolved  about 
TQ  and  the  diagram  drawn  to  scale  as  shown  in  Fig.  15  id. 

The  reactions  and  resisting  moments  will  be  determined  for 
the  fixed  beam  in  Fig.  1520. 


246 

From  statics 

or 


THEORY    OF    FRAMED    STRUCTURES 
640 


CHAP.  VI 


24 


X9X15  =  3>6oo 


Mi  =  -(640  X  9)  +  24^2  + 
5>76o  +  M  i  —  M2 


-2250 


The  angle  between  the  tangents  through  A  and  5  is 

0  =  o 

(Mi  +  ^2)24      3,600  X  24  _ 
or  —  o 

2  2 

The  tangential  deviation  at  B  is 

£2  =  o 
3,600  X  15  X  io   ,   3,600  X  9  X  18    , 


whence 


24M2  X  8          24Mi  X   16 

2  2 

=    -2,250 
=    -1,350 
=  437-5 
=   2O2.5 


SEC.  I 


RESTRAINED   AND    CONTINUOUS   BEAMS 


247 


These  results  may  be  checked  by  the  formulas  of  the  preceding 
problem.  The  M-diagram  is  drawn  to  scale  in  Fig.  152^. 

157.  Restraint  at  Both  Ends— Uniform  Load. — The  fixed 
beam  in  Fig.  153  supports  a  total  load  W,  uniformly  distrib- 
uted. Since  the  loading  is  symmetrical,  the  resisting  moment 
and  reactions  at  B  are  the  same  as  at  A .  The  area  TSQ  is  the 


!A"^> 
""l 


FIG.  153. 

M-diagram  for  a  simply  supported  beam  and  the  area  TUPQ 
represents  the  resisting  moment  M  at  each  end.  The  angle  <f> 
between  the  tangents  through  A  and  B  is  zero,  consequently 
the  area  of  the  M-diagram  is  zero,  therefore 

Ml  -f  \~jr)(2l)  =  ° 

*-_E! 

12 

The  bending  moment  at  the  center  is 

Wl  _  Wl  =  Wl 
8         12        24 

and  the  M-diagram  may  be  drawn  to  scale  as  shown. 


248 


THEORY    OF   FRAMED    STRUCTURES 


SEC.  II.     CONTINUOUS  BEAMS 


CHAP.  VI 


158.  Continuous  beams  rest  on  more  than  two  supports 
and  have  more  than  one  span.  Consequently  they  are  stati- 
cally indeterminate,  and  one  or  more  elastic  equations  are 
necessary  for  finding  the  reactions.  An  introduction  to  the 


problem  will  be  made  in  connection  with  the  beam  in  Fig.  154, 
which  supports  a  load  of  10  Ib.  at  D.  The  tangent  FG  is  drawn 
to  the  elastic  curve  through  C  at  the  middle  of  the  span.  The 
product  El  of  the  modulus  of  elasticity,  and  the  moment  of 
inertia  is  assumed  as  unity,  and  the  weight  of  the  beam  is  not 
considered.  The  M-diagram  is  PQV.  When  the  beam  is 
supported  at  A  and  B  only,  the  reactions  R,  the  bending  mo- 
ments M ,  the  tangential  deviations  t,  and  the  deflections  A  are 
as  tabulated  in  the  column  i  of  Table  I.  Now  suppose  that  an 


SEC.  II 


RESTRAINED    AND    CONTINUOUS   BEAMS 


249 


upward  force  Rc  =  i  Ib.  is  applied  at  C.  The  shape  of  the 
elastic  curve  will  be  altered  in  accordance  with  the  data  given 
in  column  2.  The  moments  are  statically  determinate;  and 
if  a  new  M-diagram  is  drawn  to  scale  an  angle  will  appear  at  S, 


FIG.  756 

the  line  QV  no  longer  remaining  straight.  The  quantities  / 
and  A  (column  2)  are  computed  from  this  diagram.  Columns  3, 
4,  5  and  6  give  similar  data,  as  the  upward  force  at  C  is  in- 
creased by  i-lb.  increments.  The  table  clearly  shows  that  for 
each  increase  of  i  Ib.  in  Rc,  Ac  is  decreased  4,500  units.  The 
original  deflection  Ac  was  (from  column  i)  25,560  units;  and 
it  is  clear  that  if  the  point  C  is  raised  4,500  units  by  each  in- 


250 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  VI 


crease  of  i  Ib.  in  Rc,  the  force  necessary  to  raise  the  point  C 
to  a  level  with  A  and  B  is 

Rc  _ 25,560  681b< 

4,5°° 

By  balancing  the  moments  about  either  A  or  B,  the  end  reac- 
tions, RI  and  R±,  may  be  determined  as  given  in  column  7. 

A  very  interesting  study  of  the  successive  stages  in  the 
transformation  of  the  elastic  curve  may  be  made,  if  diagrams 
similar  to  Fig.  154  are  drawn  to  scale  in  accordance  with  the 
data  given  in  each  column  of  the  table.  These  curves  have 
been  combined  in  Fig.  155,  and  are  numbered  to  correspond 
with  the  columns  in  the  table.  When  Rc  =  4  Ib.,  RI  =  6  Ib., 
and  ^4  =  0;  consequently  Mc  =  Mz  =  o  and  the  M-diagram 
is  PQS  (Fig.  154),  the  point  S  coinciding  with  U.  Since  the 
area-moment  of  VSU  about  V  is  zero,  /4  =  o,  G  coincides  with 
B ;  and  the  elastic  curve  CB  being  bent  by  no  moment  becomes 
a  straight  line  coinciding  with  the  tangent  CG\  (see  curve  5). 
When  Rc  =  5  Ib.;  R^  being  negative,  acts  downward;  Mc  and 
M 3  become  negative  moments;  and  /4,  being  negative  for  curve 
6,  is  measured  above  B  instead  of  below.  Finally,  when  Rc  = 
5.68  Ib.,  and  the  point  C  has  been  raised  to  the  line  AB  (curve 
7);  the  tangent  through  C  makes  equal  intercepts  on  the 

TABLE  I 


i 

2 

3 

4 

5 

6 

7 

Rc  =  0 

Rc  =  i 

Rc   =    2 

Rc  =  3 

Re  =  4 

Rc  =  5 

Rc  =  5.68 

Rl 

8 

7-5 

7-0 

6-5 

6.0 

5-5 

5-i6 

R* 

2 

i-5 

I  .O 

0-5 

0.0 

-5-o 

—0.84 

M2 

96 

90 

84 

78 

72 

66 

61  .92 

Mc 

60 

45 

30 

15 

o 

-i5 

-25.2 

Mz 

30 

22.5 

15 

7-5 

o 

-7-5 

—  12.6 

/i 

33,120 

28,620 

24,120 

19,620 

15,120 

10,620 

7,56o 

t. 

ii  ,664 

9,720 

7,776 

5,832 

3,888 

i,944 

622.08 

h 

5,625 

4,218.75 

2.,  8l2.'5 

1,406.25 

o 

—  1,406.25 

-2,362.5 

/4 

18,000 

13,500 

9,OOO 

4,5oo 

o 

-4,50° 

-7,56o. 

A2 

18,432 

15,876 

13,320 

10,764 

8,208 

5,652 

3,9i3-93 

Ac 

25,560 

21  ,O6O 

16,560 

I  2  ,  060 

7,560 

3,060 

o 

A3 

16,155 

13,061.25 

9,967-5 

6,873.75 

3,780 

686.25 

-1,417-5 

SEC.  II  RESTRAINED   AND    CONTINUOUS   BEAMS 

ordinates  through  A  and  B,  the  one  below  and  the  other  above 
AB,  hence  h  =  —  /4. 

Let  d2,  dc  and  d3  represent  respectively  the  differences  in 
A2,  Ac  and  A3,  for  consecutive  columns  i  to  6  in  Table  I; 
then  dz  =  2,556;  dc  =  4,500  and  dz  =  3,093.75.  These  dif- 
ferences in  deflection  between  any  two  curves  i  to  6  (Fig.  155) 
are  caused  by  a  difference  of  i  Ib.  in  the  force  Rc,  irrespective 
of  the  magnitudes  RI,  Rc  or  R*.  The  fact  that  these  differences 
are  constant  for  any  ordinate  whether  2-2,  C-C,  3-3  or  any 
other  ordinate  between  A  and  B  which  might  be  chosen,  is  due 
to  the  constant  differences  in  Mz,  Mc  and  M3.  Let  m^,  mc 
and  w3  represent  these  differences;  then  from  Table  I,  nh  =  6 
mc  =  15  and  ms  =  7.5.  By  reference  to  Fig.  156  it  is  at  once, 
apparent  that  nh,  mc  and  w3  represent  the  corresponding 
ordinates  in  the  M-diagram  for  the  beam  in  question;  when 
supporting  a  single  load  of  i  Ib.  at  the  center.  If  El  is  again 
taken  as  unity,  then 

dc  =  h  =  15  X  15  X  20  =  4,500 

/2  =  (6  X  9  X  6)  +  (15  X  9  X  12)  =  1,944 

*3  =  (7-5  X  7-5  X  5)  +  (15  X  7-5  X  10)  =  1,406.25 

dz  =  h  —  h  =  2,556 

dz  =  h  -  /8  =  3,093.75 

Consequently  the  difference  d  between  any  two  curves  from  i  to 
6  (Fig.  155)  on  any  ordinate,  caused  by  a  difference  of  i  Ib.  in 
Rc,  equals  the  deflection  at  the  same  ordinate  caused  by  a  load 
of  i  Ib.  at  C  (Fig.  156). 

Hence,  if  a  beam  is  continuous  over  three  supports,  the 
intermediate  reaction  Rc  at  C  (not  necessarily  at  the  center) 
may  be  determined  as  follows:  Remove  the  intermediate 
reaction;  find  the  end  reactions  as  for  a  simple  beam  and  com- 
pute the  deflection  A  at  C;  finally,  compute  the  deflection  d  at  C 
due  to  a  load  of  i  Ib.  at  C.  Then 


159.  Application  of  Maxwell's  Theorem.  —  Maxwell's  theorem 
of  reciprocal  displacements  may  be  used  in  finding  the  reactions 
of  a  continuous  beam  on  three  supports.  The  deflection  at  D 


252  THEORY    OF   FRAMED    STRUCTURES  CHAP.  VI 

(Fig.  156)  caused  by  a  load  of  i  Ib.  at  C  is  J2  =  2,556;  hence,  the 
deflection  at  C  caused  by  a  load  of  i  Ib.  at  D  is  2,556;  and  the 
deflection  at  C  caused  by  10  Ib.  at  D  is  Ac  ==  2,556  X  10  = 
25,560,  as  shown  in  Fig.  154.  Since  the  deflection  at  C,  caused 
by  a  force  of  i  Ib.  at  C,  is  dc  =  4,500;  then  the  force  at  C 
necessary  to  raise  the  point  C  to  a  level  with  AB  is 

Rc  =  ip^  =  25,560  =  ^ 

dc          4,500 

Hence,  if  a  beam  is  continuous  over  three  supports  at  A,C 
and  B,  and  supports  loads  PI,  P2  and  P3  at  any  points  i,  2  and 
3,  the  intermediate  reaction  Rc  may  be  found  as  follows: 
Remove  the  loads  P  and  the  reaction  Rc:  place  i  Ib.  at  C,  and 
compute  the  deflections  AI,  A2,  AS  and  Ac  at  the  points  i,  2,  3 
and  C.  Then 

P3A3 


c          -  —  - 

Ac 

The  reactions  at  ^4  and  B  may  be  determined  by  statics  after 
the  reaction  at  C  is  known.  It  was  not  necessary  to  apply 
Maxwell's  theorem  to  the  intermediate  reaction,  for  either 
end  reaction  might  have  been  determined  equally  as  well  by 
this  theorem. 

The  ratio  of  d2  to  dc  (Fig.  156)  may  be  determined  from 
Table  I,  page  222.  The  load  i  Ib.  is  at  the  center,  hence 
k  =  0.5.  When  c  =  0.2,  F  =  0.071;  when  c  =  0.5,  F  =  0.125; 
hence 

d%  _  0.071 
dc       0.125 


and  Rc  =  =  5.68  Ib. 

dc 

160.  The  Conventional  Method.  —  The  method  which  is 
generally  employed  in  finding  the  reactions,  by  area-moments, 
of  a  beam  continuous  over  three  supports,  will  be  given  in  con- 
nection with  Fig.  157.  The  elastic  curve  is  not  shown  and  no 
speculation  with  reference  to  its  form  will  be  made.  Let  FG 
represent  the  tangent  to  the  elastic  curve  drawn  through  C. 
Its  slope  is  unknown;  it  may  be  positive  or  negative.  One 
thing  is  certain.  Since  the  two  spans  are  equal  in  length,  the 


SEC.  II 


RESTRAINED   AND    CONTINUOUS  BEAMS 


253 


tangential  deviations  /i  and  /2  are  equal  in  magnitude.     They 
have  opposite  signs,  since  one  is  measured  above  the  line  AB 


and  the  other  below  it. 
Hence 


254  THEORY    OF    FRAMED    STRUCTURES  CHAP.  VI 

All  reactions  will  be  assumed  to  act  upward,  hence  a  negative 
numerical  value  for  the  solution  of  a  reaction  indicates  that  its 
action  is  downward. 

The  bending  moment  is  known  only  at  two  points  A  and  B. 
The  M-diagram  may  be  constructed  in  several  ways. 

In  Fig.  157^,  all  ordinates  are  expressed  in  terms  of  the  end 
reactions. 

£//i  =  12^1(6  X  8)  +  12^1(9  X  18)  +  30^3(9  X  24) 
£//2  =  30^3(15  X  20) 

h  =  -h 

whence  jRi  =  —43^3 

From  statics  3<xRi.—  180  =  30^3 

Ri  =  R*  +  6 

Therefore  Ri  =  5.16 

R2  =  5.68 

^3   =    -0.84 

In  Fig.  157^,  the  M-diagrams  for  the  load  at  D  and  for  the 
reaction  at  C  are  sketched  separately.  PQV  is  the  M-diagram 
when  the  center  reaction  is  removed  and  AB  considered  as  a 
simple  beam,  supporting  the  load  at  D.  PUV  is  the  M- 
diagram  when  the  load  at  D  is  removed,  and  AB  considered  as 
a  simple  beam  held  in  equilibrium  by  forces  at  A,  B  and  C. 
Elti  =  (96  X  6  X  8)  +  (96  X  9  X  1 8)  +  (60  X  9  X  24)  + 

M(is  X  20) 
£7/2  =  (60  X  15  X  20)  +  M(i$  X  20) 

h  =  -/2 
whence 

M  =  -85.2 

The  bending  moment  at  C  is 

M  =  SU=6o-M=  -25.2 
hence  •    —25.2  =  $oRi  —  180  =  $oRz 

Therefore  #1  =  5.16 

R2  =  5-68 
Rs  =  —  0.84 

In  Fig.  157*2,  the  M-diagram  is  drawn  by  first  considering 
that  AC  and  CB  are  simple  beams;  i.e.,  by  assuming  no  con- 
tinuity and  no  bending  moment  at  C;  thus  PQS  is  the  M- 


SEC.  II 


RESTRAINED   AND    CONTINUOUS   BEAMS 


255 


diagram  for  the  simple  beam  AC  supporting  10  Ib.  at  D. 
There  is  no  corresponding  diagram  for  CBj  since  there  is  no 
load  in  that  span.  The  area  PUV  is  then  added  to  provide  for 
the  bending  moment  on  account  of  continuity  at  C. 


3000 ] 


I000# 


1000' 


___>< £_'__><___£'___>. 


(b) 

FIG.  158. 

=  (72  X  6  X  8)  +  (72  X  9  X  18)  +  M(i$  X  20) 
=  M(i$  X  20) 
ti  =  -tz 
whence 

M  *=  —25.2  as  before. 

After  the  reactions  have  been  determined,  the  bending 
moments  at  C  and  D,  and  the  deflection  at  any  point  may  be 
computed.  The  elastic  curve  when  drawn  will  have  the  general 
configuration  shown  in  Fig.  159. 


256 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  VI 


161.  In  Fig.  1580,  PQS  is  the  M-diagram  when  AC  is  con- 
sidered as  a  simple  beam,  and  SOTV  is  the  M-diagram  when 
CB  is  considered  as  a  simple  beam;  then  the  area  PUV  is  added 
to  provide  the  bending  moment  on  account  of  the  continuity. 
Let  the  tangent  to  the  elastic  curve  be  drawn  through  C  and 
let  t\  and  fa  represent  the  tangential  deviations  at  A  and  B 


FIG.  159. 

respectively,  then 

Elh  =  (9,000  X  6  X  6)  +  M(6  X  8) 

EItz  =  (6,000  X  3  X  4)  +  (6,000  X  6  X  9)  +  (6,000  X   3   X 

14)  +  M(g  X  12) 
t\:  —fa::  12:18 
whence 
M  =  —6,300 
The  M-diagram  may  now  be  drawn  to  scale  as  shown  in  Fig. 


—6,300  =  i2Ri  —  18,000 

*i  =  975 
—  6,300  =  i&Rs  —  6,000  —  12,000 

RZ  =  650 

Ri  =  5,000  -  975  -  650  =  3,375 


From  statics 


SEC.  II  RESTRAINED   AND   CONTINUOUS  BEAMS  257 

162.  The  general  expressions  for  RI,  R2  and  R$  will  now  be 
developed  in  connection  with  Fig.  159.  Let  the  tangent  to  the 
elastic  curve  be  drawn  through  C  and  let  t\  and  /2  represent  the 
tangential  deviations  at  A  and  B  respectively;  then 

=  Pk(i  -  k)l 


/1/2    =    ~ 


_  - 

2(/i  +  fe) 


Since  ^  is  less  than  unity,  k  —  k3  is  positive;  hence  M  is  a 
negative  bending  moment,  and  Rs  is  a  negative  reaction  acting 
downward. 

When  the  two  spans  are  of  equal  length  /,  the  reactions  are 

*i  =  -(#  -  5k  +  4)  (9) 

4 
£2    =^(_2£3    +    6£)  (IQ) 

4 

R*  =  ~(k*-k)  (n) 

4 

163.  In  Fig.  i6oa  the  parabola  PQS  is  the  M-diagram,  when 
AC  is  considered  as  a  simple  beam;  and  the  parabola  STV  is  the 
M-diagram  when  CB  is  considered  as  a  simple  beam.  The 
triangle  PUV  is  added  to  represent  the  bending  moment  on 

account  of  the  continuity.     If  the  tangent  to  the  elastic  curve 
17 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  VI 


be  drawn  through  C,   and  /i  and  /2  represent  the  tangential 
deviations  at  A  and  B  respectively;  then 


POOO  Ibs.per  ft 


E//I  =  (18,000  X  12  X  -  X  6)  +  M(6  X  8) 

O 

=  864,000  +  48M 
Elh  =  (81,000  X  18  X  -  X  9)  +  M(g  X  12) 

o 

=  8,748,000  +  io8M 
3/1  =  —  2/2 
M  =  -  55,800 


SEC.  II 


RESTRAINED   AND    CONTINUOUS  BEAMS 


259 


The  M-diagram  may  now  be  drawn  to  scale  as  shown  in  Fig. 
1606. 
From  statics      —55,800  =  i2Ri  —  (12,000  X  6)  =  i&R3  — 

(36,000  X  9) 


&  =  31,75° 

RS  =  14,900 

164.  Two  Unequal  Spans,   Supporting  Unequal  Uniform 
Loads.  —  The  general  expressions  for  Ri,  R%  and  Rs  for  a  con- 


w, 


FIG.   161. 

tinuous  beam  of  two  unequal  spans  l\  and  k,  supporting  unequal 
uniform  loads,  w\  and  w%  per  unit  of  length,  will  now  be  de- 
veloped in  connection  with  Fig.  161.  The  M-diagram  is 
drawn  as  in  the  preceding  problem.  If  the  tangent  to  the 
elastic  curve  is  drawn  through  C;  and  t\  and  h  represent  the 
tangential  deviations  at  A  and  B  respectively;  then 


Will4   .   Mli' 

—    I      

24  3 


24 


M  =  - 


(12) 


260 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  VI 


When  the  spans  are  equal  in  length  /  and  the  uniform  load  w 
per  unit  of  length  is  the  same  in  both  spans,  Eq.  (12)  reduces  to 

*--f  (13) 

165.  When  a  continuous  beam  supports  a  combination  of 
uniform  and  concentrated  loads,  it  will  be  found  expedient  to 
sketch  the  M-diagram  in  parts  as  shown  in  Fig.  162.  The 
portion  (a)  is  the  M-diagram  for  the  concentrated  loads  when 
no  continuity  is  considered  at  C;  and  the  portion  (b)  is  a  similar 


diagram  for  the  uniform  loads.  The  continuity  is  provided  for 
by  the  portion  (c).  If  the  tangent  to  the  elastic  curve  is 
drawn  through  C;  and  /i  and  k  represent  the  tangential  devia- 
tions at  A  and  B]  then 


=  (9,000  X  6  X  6)  +  (18,000  X  12  X  -  X  6)  + 

M(6  X  8)  =  1,188,000  +  48M 

Elk  =  (6,000  X  3  X  4)  +  (6,000  X  6  X  9)  + 

(6,000  X  3  X  14)  +  (81,000  X  18  X  -  X  9)  + 

M(g  X  12)  =  9,396,000  +  io8M 
3/1  =  —2/2 
M  =  —62,100 


SEC.  II  RESTRAINED   AND    CONTINUOUS   BEAMS  261 

The  reactions  may  now  be  determined  by  the  principles  of 
statics. 

The  value  of  M  may  also  be  determined  from  Eqs.  (5)  and  (12). 
Eq.  (5)  is  applicable  to  the  concentrated  loads.  For  the 
load  at  Z>;  P  =  3,000,  k  =  %,  h  =  12  and  h  =  18;  hence 


M= 


For  the  load  at  F\  P  =  1,000,  k  =  -,    l\  =  18    and   h  =  12; 

o 

hence 

1,000  X  324(-  -  - 
* 


2(12  +  1  8) 
e  load  at  E;  P  = 
hence 


= 


2 

For  the  load  at  E;  P  =  1,000,  k  =  -,  /i  =  18   and  /2  =  12, 

o 


1,000  X  324( 

M  = 


EN 

2(12  +  18) 

Hence  the  bending  moment  at  C,  due  to  the  three  concentrated 
loads, 

M  =  —2,700  —i,  600  —2,000  =  —6,300 

which  agrees  with  bending  moment  at  C  for  the  beam  in  Fig. 

158. 

Equation   (12)   is  applicable  to  the  uniform  loads,   where 
Wi  =  1,000,  wz  =  2,000,  li  =  12  and  li  =  18;  hence 


which  agrees  with  the  bending  moment  at  C  for  the  beam  in 
Fig.  1  60.  The  total  bending  moment  at  C  for  the  combined  uni- 
form and  concentrated  loads  is 

M  =  —6,300  —  55,800  =  —62,100 

as  previously  determined. 

166.  The  continuous  beam  in  Fig.  163  supports  a  uniform 
load  of  1,000  Ib.  per  foot  over  a  part  of  the  span  AC.  The  area 
PQSTW  is  the  M-diagram,  when  AC  is  considered  as  a  simple 


262 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VI 


span.  Let  the  tangent  to  the  elastic  curve  be  drawn  through 
C,  and  let  t\  and  /2  represent  the  tangential  deviations  at  A 
and  B  respectively.  In  finding  the  area-moment  of  PQSTW 
about  P,  the  parabolic  area  QST  is  encountered.  This  area 
has  all  the  properties  of  the  area  Q'S'T1,  which  is  the  M-diagram 
for  a  simple  beam  12  ft.  long  supporting  a  uniform  load  of  1,000 

I000*/l.f. 


--><-- 


PIG.  163. 

lb.  per  foot  over  its  entire  length;  hence  the  area-moment  of 
PQSTW  about  P  may  be  found  as  follows: 

area  PQN        43,200  X  3    X   4  =      518,400 

43,200  X  6    X  10  =  2,592,000 

57,600  X  6    X  14  =  4,838,400 

57,600  X  6    X  22  =  7,603,200 


area  QNO 
area  QTO 
area  TOW 
area  QST 


18,000  X   12  X  %  X    12   =    1,728,000 


17,280,000 

=  17,280,000  +  M(i$  X  20) 
£//2  =  Jlf  (30  X  40) 

2/i=    ~/2 

M  =  —19,200 

The  reactions  are  statically  determinate  when  M  is  known. 
The  value  of  M  may  also  be  determined  by  the  use  of  Eq.  (5) 


SEC.  II 


RESTRAINED   AND    CONTINUOUS   BEAMS 


263 


in  which  P  is  a  concentrated  load  at  the  distance  kl\  from  A. 
In  the  present  case  let  P  represent  the  weight  of  an  element,  of 
length  dkli  at  the  distance  kli  from  A  ,  then 
P  —  i  ,000  lidk 


whence 


„, 

dM  = 


.  . 

2  (/i  +  k) 

The  value  of  M  may  be  found  by  integrating  between  the 
limits  k  =  0.2  and  k  =  0.6,  hence 


fk2  £410- 

=      -150,000      - 

12  4  Jo. 


=  —19,200 

167.  The  beam  in  Fig.  164  is  continuous  over  four  supports. 
Two  elastic  equations  are  required,  in  addition  to  the  two 


FIG.  164. 

static  equations  which  may  be  written,  for  the  determination  of 
Rij  Rt,  Rs  and  R^  Let  h  and  /3  represent  the  tangential 
deviations  at  A  and  Z>,  for  the  tangent  to  the  elastic  curve  at 
C;  and  let  h  and  /4  represent  the  tangential  deviations  at  C 
and  B,  for  the  tangent  to  the  elastic  curve  at  D\  then 

ati  =  —  /3 

and  /2  =  — 0/4 

PQW  is  the  M-diagram  when  AC  is  considered  as  a  simple 


264  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VI 

span,  to  which  the  diagram  PUSV  is  added  to  provide  for 
continuity. 

Elh  =  Pk(i  -  k)l(j  l\-\kl  +  ±(i  -  «)/j  +  M,(~2  ;V*  l\ 


6  v  3 

£7/3  =  Mi(-  al}(-  al}  +  W-  al}(-  al} 
\2     /\3     /  \2     /\3     / 

=   -— (2Afi  +  Mz) 


Elh  =  M       o/        al 


-  al }  I  -  al } 


Whence   K ,  = 


3^  +  Sa  +  4 

PJ(, 

Mz  = 


W 

—  k) 


4 

~  ^8)("  + 


#2  +  8a  +  4 

P(^-^)(2a2  +  5^  + 
"  (3a2  +  Sa  +  4)a 


R*  ==  — o  .   o — r~ 
3^     i    o^  ~r4 

168.  Three  Equal  Spans — Uniform  Load. — The  beam  in  Fig. 
165  is  continuous  over  three  equal  spans  /,  and  supports  a  uni- 
form load  of  w  Ib.  per  unit  of  length.  Since  the  beam  is  sym- 
metrical about  the  center,  the  ordinates  in  the  M-diagram  at 
B  and  C  are  equal;  and  only  one  elastic  equation  is  necessary. 
Let  the  tangent  to  the  elastic  curve  be  drawn  through  B,  and 


SEC.  II 


RESTRAINED    AND    CONTINUOUS   BEAMS 


265 


let  t\  and  /a  represent  the  tangential  deviations  at  A    and  C 
respectively;  then 


wl*      Ml2 
24         3 


FIG.  165. 


__  wl*      Ml2 

24  2 


24          3 


24 


M  =  - 


10 


10 


IO 


If   the   M-diagram   had   not  been   symmetrical,   either   on 
account  of  unsymmetrical  loading  or  variation  in  span  lengths, 


i 

266  THEORY   OF   FRAMED   STRUCTURES        .          CHAP.  VI 

or  both;  the  ordinates  at  B  and  C  could  not  have  been  assumed 
equal.  In  this  case  there  would  have  been  two  unknown 
ordinates,  Mz  at  B  and  If  3  at  C;  and  two  elastic  equations  would 
have  been  necessary  for  a  solution. 

These  two  equations  may  be  written  by  establishing  a  rela- 
tion between  /i  and  /3,  and  a  relation  between  /i  and  /4;  or  one 
tangent  may  be  drawn  through  B  and  another  through  C. 

169.  Four  Equal  Spans  —  Uniform  Load.  —  The  beam  in  Fig. 
1  66  is  continuous  over  four  equal  spans  /,  and  supports  a 
uniform  load  of  w  Ib.  per  unit  of  length.  Since  the  beam  is 
symmetrical  about  the  center,  there  are  only  two  unknown 
ordinates  Mi  and  M  3  to  be  determined,  for  which  two  elastic 
equations  are  necessary.  On  account  of  symmetry,  the  tangent 
to  the  elastic  curve  through  C  is  horizontal,  consequently  the 
tangential  deviations  t\  at  A  and  tz  at  B  are  zero;  hence 


or  6Mz  +  sMz  =  —  wl 


25 


Whence  Rl  =  R,=wl 

25 

R*  =  #3  =  ^|w/ 

25 

If  for  any  reason  the  M-diagram  had  not  been  symmetrical, 
the  ordinates  at  B  and  D  could  not  have  been  assumed  equal; 
and  the  tangent  to  the  elastic  curve  through  C  would  not  have 
been  horizontal.  There  would  have  been  three  unknown 
ordinates,  M2  at  B,  Ms  at  C  and  M4  at  D;  and  three  elastic 
equations  would  have  been  necessary  for  a  solution.  These 
three  equations  may  be  written  as  follows  :  (a)  draw  a  tangent 


SEC.  II 


RESTRAINED   AND    CONTINUOUS  BEAMS 


267 


to  the  elastic  curve  at  B  and  establish  a  relation  between  the 
tangential  deviations  at  A  and  C;  (6)  draw  a  tangent  to  the 
elastic  curve  at  C  and  establish  a  relation  between  the  tangen- 
tial deviations  at  B  and  D;  (c)  draw  a  tangent  to  the  elastic 
curve  at  D  and  establish  a  relation  between  the  tangential 
deviations  at  C  and  E. 

The  three  equations  may  also  be  written  as  follows: 
Let  the  tangent  to  the  elastic  curve  be  drawn  through  B; 
and  let  t\,  /3,  /4  and  /5  represent  the  tangential  deviations  at 
A}   C,  D  and  E  respectively.     Then  establish  a  relation  be- 


ft1 


FIG.   166. 


tween  /i  and  /3,  a  second  relation  between  t\  and  £4,  and  a  third 
relation  between  t\  and  /5. 

In  case  the  beam  is  restrained  at  each  end  by  being  fixed 
in  a  wall;  the  M-diagram  presents  five  unknown  ordinates,  or 
one  at  each  point  of  support.  This  condition  requires  two 
additional  elastic  equations,  or  five  in  all.  These  two  equa- 
tions may  easily  be  written,  since  the  tangents  to  the  elastic 
curve  through  A  and  E  are  horizontal. 

170.  Coefficients  for  Pier  Reactions. — When  the  spans  are 
equal  in  length  and  the  load  is  uniform  throughout,  the  reaction 
at  each  support  may  be  found  by  multiplying  the  total  load 
on  each  span  by  the  coefficients,  as  given  in  the  following 
table.  The  Roman  numerals  represent  the  number  of  spans 
over  which  the  beam  is  continuous. 


268  THEORY    OF   FRAMED    STRUCTURES  CHAP.  VI 

REACTION  COEFFICIENTS 

I    !-! 


HI        ___. 

IO         10         10         10 

IV     H_3^_^>_3^_!i 
28      28      28      28      28 

V     15_43  _37  _37  _43  _  £5 
38      38      38      38      38      38 

SEC.  III.     CLAPEYRON'S    THEOREM    OF    THREE    MOMENTS 

171.  Clapeyron's  theorem  may  be  used  to  establish  a  rela- 
tion between  the  bending  moments  at  any  three  consecutive 
supports  of  a  beam  of  uniform  cross-section,  as  shown  in  Fig. 
167.  Let  li  and  /2  represent  the  lengths  of  any  two  adjacent 
spans,  which  support  the  uniform  loads  of  w\  and  w2  per  unit  of 
length  respectively;  and  let  MQ,  MI  and  M2  represent  the  bend- 
ing moments  at  the  three  supports.  Let  the  tangent  to  the 
elastic  curve  be  drawn  through  the  middle  support;  and  let  /o 
and  t<i  represent  the  tangential  deviations  at  the  left  and  right 
supports  respectively;  then 


1    7        _l_    \f 
-/I  1+  Mi     - 

2 


1   1  V2  7   \ 
/I        -/l  ) 

2    /\3    / 


»  £.  * 

24  6 


8 

,  , 


1  ,  V? 

2  /\3 


24 

/0/2    =    —  fe/1 

Whence 


i  +  fe) 

4 

In  any  continuous  beam,  not  restrained  at  the  ends  having 
n  spans,  there  are  n-i  unknown  bending  moments  at  the 
intermediate  supports,  which  may  be  determined  by  writing 


SEC.  Ill 


RESTRAINED   AND    CONTINUOUS   BEAMS 


269 


n-i  equations  similar  to  (14).  Equations  of  this  type  are 
applicable  when  any  span,  taken  at  random,  supports  either  a 
uniform  load  over  its  entire  length,  or  no  load  whatever. 
When  two  adjacent  spans  have  the  same  length  /,  and  support 
the  same  load  w  per  unit  of  length,  Eq.  (14)  reduces  to 


An  application  of  Eq.  (14)  will  be  made  in  connection  with 


PIG.  167. 


1000%.  I.  f 


immmiimiiiimmiii 
|*.-0£-4t 20- >H tf-'--- 


5' 


FIG.   168. 


the  continuous  beam  of  uniform  cross-section,   having    five 
spans  (Fig.  168).     First  and  second  spans: 

=  —  (o  +  1,000  X  2o3) 
4 


ioMo  +  2Afi(io  +  20)  + 
Second  and  third  spans: 

2oMi  +  2M2(20  +   15)   + 

Third  and  fourth  spans: 

15^2  +  2^3(15  +  10)  +  ioM4 

Fourth  and  fifth  spans: 

ioM3  +  2^1/4(10  +  15)  +  15^5  =  —  -[(500  X  io3)  + 

(1,500  X  is8)! 


—    —-(l,000  X    203 


—  I(o  +  500  X  io3) 
4 


270 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VI 


Since  the  beam  is  simply  supported  at  each  end, 

Mo  =  O 

Mb  =  o 

A  solution  of  these  equations  will  give  the  bending  moments 
at  the  points  of  support,  and  the  reactions  may  be  determined 
from  the  principles  of  statics. 

SEC.  IV.     PARTIALLY  CONTINUOUS  BEAMS 

172.  No  Shear  Transmitted. — It  is  frequently  desirable 
to  consider  a  structure  in  which  the  continuity  is  imperfect. 
A  swing  truss  bridge  on  four  supports,  designed  with  parallel 


-----x---c*l--->< 1 


w 


H 


U  5 

FIG.   i6ga. 

chords  and  very  light  web  members  in  the  center  span,  so 
that  no  shear  can  be  transmitted  between  the  two  inside 
supports  is  a  structure  of  this  kind.  Such  structures  are 
called  partially  continuous,  and  their  treatment  will  be  illus- 
trated by  the  beam  in  Fig.  1690. 

It  is  assumed  that  bending  moment,  but  no  shear  exists  in 
the  center  span;  hence  Rz  =  —  R^  and  the  bending  moment  M 
at  B  equals  the  bending  moment  at  C.  Since  the  continuity 
of  the  beam  is  broken  at  B  and  C,  the  elastic  curve  is  not  con- 
tinuous, but  forms  cusps  at  these  points;  and  the  tangent  FG 


SEC.  IV  RESTRAINED   AND    CONTINUOUS   BEAMS  271 

to  the  elastic  curve  for  AB  at  B  is  not  tangent  to  the  elastic 
curve  for  BC.  Similarly  the  tangent  HG  to  the  elastic  curve 
for  CD  at  C  is  not  tangent  to  the  elastic  curve  for  BC.  Let  0i 
=  angle  ABF,  and  04  =  angle  DCH,  then 

area  WTSU 

6l  +  04  =  </>  =  -     —  JTT— 
xii 

The  tangential  deviations  at  ^4  and  Z),  being  represented  as 
measured  above  the  axis  of  the  beam,  are  considered  negative. 

-/i  =  Oil 


£70  =  Jlf  a/ 

i  =  Pk(i  -  k)l-l-kl  +  -(i  - 

2 


3 
Since  £7/i  +  EIt±  =  —EI<t>l 


_  P/3,L  Ml2   . 

Then  —  (y^  —  &3)  +  -  -  =  —  Mai2 

6  33 


4         a 
Therefore  i?i  =  P(i  -  k)  - 


4  +  oa 
P(k  -  k*) 


4  4-  6a 
P(k  - 


4  +  6a 

173.  No  Moment  Transmitted.—  The  span  in  Fig.  1696  con- 
sists of  two  restrained  beams,  connected  at  mid-span  in  such  a 
way  that  shear,  but  no  bending  moment,  can  be  transmitted 
from  one  beam  to  the  other.  The  span  therefore  represents  a 
different  phase  of  partial  continuity  from  that  of  the  previous 
problem.  The  principle  here  involved  is  employed  in  the 


272 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VI 


design  of  a  bascule  span  composed  of  two  leaves  connected  by  a 
shear  lock.  The  principle  must  be  modified,  however,  in  its 
application  to  a  bascule  span;  for  the  leaves  do  not  as  a  rule 
have  a  constant  moment  of  inertia,  nor  are  they  in  perfect 
restraint  at  the  points  of  support. 

A  constant  moment  of  inertia  and  perfect  restraint  will  be 
assumed  in  finding  the  shear  V  on  the  pin-connection  at  C, 


-Ml 


FIG.  1696. 

when  the  beam  CB  supports  the  load  P  as  shown.  The  M- 
diagram  may  be  drawn  very  easily  when  the  partially  continu- 
ous beam  ACB  is  considered  as  two  restrained  beams  sketched 
separately;  with  the  shear  at  C  considered  as  a  force  V,  acting 
upward  on  CB  and  downward  on  CA.  The  bending  moment 
at  C  is  zero.  The  M -diagram  for  CB  is  best  sketched  in  two 
parts — the  area  QST  representing  the  bending  moment  of  V, 
and  the  area  TUW  representing  the  bending  moment  of  P. 


SEC.  V  RESTRAINED   AND    CONTINUOUS   BEAMS  273 

Since  the  continuity  is  broken  at  C,  it  cannot  be  assumed  that 
the  total  area  of  the  M-diagram  is  zero,  although  the  angle  0 
between  AD  and  BD  is  zero.  The  absurdity  of  such  an  assump- 
tion is  obvious  when  k  =  i  and  the  load  P  is  at  C,  in  which  case 
it  is  clear  that  the  M-diagram  is  a  negative  area  throughout 
and  cannot  equal  zero;  neither  can  the  tangential  deviation  at 
either  A  or  B  be  equated  to  zero,  for  a  similar  reason. 

Let  /i  represent  the  tangential  deviation  for  the  beam  AC  and 
/2  for  the  beam  BC,  then 


Whence  V  =  -£2  - 


SEC.  V.     CONTINUOUS  BEAMS  IN  FOUNDATIONS 

174.  In  designing  foundations  for  high  buildings,  it  is  fre- 
quently necessary  to  place  three  columns  on  a  single  footing. 
An  attempt  is  made  to  secure  uniform  soil  pressure  by  support- 
ing the  columns  on  longitudinal  girders  resting  on  shorter  cross- 
beams. These  beams  bear  on  the  soil  and  distribute  the  column 
loads  over  a  sufficient  area  of  foundation.  In  the  design  of  such 
a  foundation  the  engineer  frequently  adopts  the  following 
course.  Having  determined  the  sum  total  of  the  three  column 
loads,  he  decides  upon  the  allowable  unit  bearing  pressure  of  the 
soil;  determines  the  required  area  of  the  foundation;  and  then 
shapes  the  footing  by  assuming  arbitrarily  either  the  length  or 
width  of  the  footing  to  suit  his  convenience.  The  last  step  in 
this  process  disregards  the  fact  that  the  girders  and  their 
loading  constitute  a  statically  indeterminate  system.  The 
arbitrary  proportioning  of  such  a  system  does  not  result  in 
uniform  soil  pressure;  hence  the  actual  soil  pressure  will  be 
greater  at  some  points;  and  less  at  other  points,  than  the 
allowable  pressure. 

Two  difficulties  are  encountered  in  making  a  rational  analysis 
of  the  problem:  (i)  The  columns  are  usually  anchored  to  the 

18 


274  THEORY   OF   FRAMED   STRUCTURES  CHAP.  VI 

girders  with  sufficient  rigidity  to  allow  for  the  transference  of 
moment.  In  the  discussion  which  follows,  this  moment  will  be 
neglected.  (2)  Since  the  girders  distort  slightly  in  performing 
their  office  of  distributing  the  loads,  the  assumption  of  uni- 
formly distributed  soil  bearing  pressure,  which  the  designer 
necessarily  makes  (in  the  present  analysis  as  well  as  in  the 
customary  procedure)  is  unavoidably  vitiated.  There  is  no 
data  available  as  to  the  importance  of  this  effect;  but  it  seems 
reasonable  to  conclude  that  a  state  of  uniform  pressure  dis- 
tribution is  more  nearly  approximated  when  the  problem  is 
solved  by  an  analysis  that  takes  account  of  the  continuity  of  the 
girders,  than  by  the  present  arbitrary  choice  of  proportions. 


600  n\ 

ToT7Sf\^ 

/<r..__,., 

AprfR-- 

70/75yH\ 

1  *>u       '     Cn 

r\Tons 

f 

i 

J. 

37.5'--- 

->k  37:5^  > 

L= 

PIG.  170. 

The  arbitrary  method  of  proportioning  will  be  given  in  connec- 
tion with  Fig.  170.  The  sum  of  the  three  column  loads  is 
3,000  tons.  If  a  permissible  soil  reaction  of  5  tons  per  square 
foot  is  assumed,  the  required  area  of  the  footing  is  600  sq.  ft. 
The  size  of  the  footing  is  taken  arbitrarily  as  75  ft.  long  and 
8  ft.  wide.  The  center  of  gravity  of  the  three  column  loads  is 
12.6  ft.  to  the  right  of  the  center  column,  which  locates  the 
center  of  the  soil  reaction;  hence  the  girders  should  extend  12.9 
ft.  beyond  the  left  column  and  20.1  ft.  beyond  the  right  column. 

A  grillage  so  designed  conforms  to  the  three  conditions  of 
static  equilibrium — i.e.  (sum  of  the  vertical  forces  equals  zero, 
sum  of  the  horizontal  forces  equals  zero,  and  the  sum  of  the 
moments  about  any  point  equals  zero).  From  the  foregoing, 
it  is  concluded  that  the  three  concentrated  column  loads 
are  supported  by  a  uniform  soil  reaction  of  5  tons  per  square 
foot  or  40  tons  per  linear  foot  of  girder.  That  this  conclusion 
is  fallacious,  however,  becomes  apparent  when  the  structure 


SEC.  V  RESTRAINED   AND   CONTINUOUS  BEAMS  275 

(Fig.  170)  is  inverted  (Fig.  171);  and  the  soil  pressure  repre- 
sented as  a  uniform  load  of  40  tons  per  linear  foot  resting  upon 
three  supports.  It  is  therefore  evident  that  the  problem  is 
statically  indeterminate,  and  the  odds  are  greatly  against  the 
probability  that  the  reactions  are  600,  900  and  1,500  tons 
respectively. 

The  rational  method  takes  account  of  the  continuity  of  the 
longitudinal  girders.  Three  independent  simultaneous  equa- 
tions are  required  for  the  determination  of  the  three  reactions 
(Fig.  171).  Two  of  these  equations  SF  =  o  and  SM  =  o  are 


FIG.  171. 

supplied  by  the  principles  of  statics.  The  third  or  elastic 
equation  is  derived  by  drawing  the  tangent  FG  through  C,  and 
establishing  the  relation  between  /i  and  k.  The  application 
of  these  three  equations  to  practical  cases  differs  in  detail, 
according  to  the  physical  conditions  governing  the  problem. 
Three  of  several  possible  cases  are  illustrated  in  Figs.  172,  173 
and  174,  and  will  be  considered  separately.  In  each  case  the 
following  points  are  to  be  observed:  Each  figure  is  shown 
inverted  for  convenience.  The  known  column  reactions  are 
represented  by  P,  Q  and  R.  The  spacing  of  the  columns,  being 
fixed  by  the  architectural  features,  is  known;  and  represented 
by  c  and  d.  The  tangential  deviations  t\  and  h,  not  shown  in 
Figs.  172,  173  and  174,  are  to  be  taken  as  represented  in  Fig. 
171.  Three  quantities,  differing  in  each  case,  are  to  be  deter- 
mined by  a  solution  of  the  three  independent  simultaneous 
equations  cited. 

175.  Case  I.  Projections  Not  Limited  by  Site. — In  Fig.  172, 
let  w  represent  the  intensity  of  the  uniform  soil  pressure  in 
pounds  per  linear  foot.  If  the  architectural  features  do  not 
limit  the  end  projections  a  and  b  of  the  main  grillage  girders; 
it  will  be  possible  to  attain  this  condition  of  uniform  soil 


276  THEORY    OF   FRAMED    STRUCTURES  CHAP.  VI 

pressure  by  selecting  a,  b  and  w  accordingly.     The  two  static 
equations  are 

P  +  Q  +  R-(a  +  b  +  c  +  d)w  =  o  (15) 

aP  +  (a  +  c)Q  +  (a  +  c  +d)  R 


-~2(a  +  b  +  c  +  d)2  =  o     (16) 

The  elastic  equation  is 

he  ,     x 

s  !  ~d  (l^) 

The  expressions  for  t\  and  ti  may  be  determined  as  follows: 


llMMUH  "Mill  II  Mill  ii  ii  iirrrm- 


iirrrm-n 
<----£  ----  J 


d 

PIG.  172. 

The  bending  moment  between  P  and  Q  at  any  distance  x  from 
Pis 

Mi  =  Px  --  w(a  -\-  x)2 
2 

hence     £//1  =    f  M^xdx  =  **  -  +  ^-  +  c- 

Jo 

e  ben 
tance  x  from  7?  is 


3         2     2  3         5 

Similarly,  the  bending  moment  between  Q  and  R  at  any  dis- 


= Rx  -  -w(b 

2 


hence     Eftt  =       M.xdx  .  Z  ±  -.Sir  L  +  i   1  +!! 

Jo  3          2\    2  3          4 

Substituting  the  values  of  t\  and  h  in  Eq.  (17)  and  reducing 

-  ^c4)  c 


Eliminating  w  from  Eqs.  (15),  (16)  and  (170) 


p_    + 

6c(P  +  Q.+  R)a*  +  6J(P  +  Q  +  £)62  +  8(c2<2 

a  +  8(J2P  +  ^Q  --  c2P)b  +  (-5C3  +  3d*  -  Sc2d)P  +  3 

?  =  o  (19) 


SEC.  V 


EESTRAINED    AND    CONTINUOUS   BEAMS 


277 


Illustrative  Problem.— Let  c  =  12  ft.,  d  =  30  ft.,  P  =  60° 
tons,  Q  =  goo  tons  and  R  =  1,500  tons. 

Substituting  the  numerical  values  in  Eqs.  (18)  and  (19),  and 
reducing 

b  =  a  +  7-2 

io02  +  256*  —  3720  +  4686  =  10,374 
whence  a  =  7.8  ft.,  and  b  =  15  ft. 

The  length  of  the  base  is  42  +  22.8  ft.  or  64.8  ft.,  and  the  soil 
pressure  per  linear  foot  is 

w  =  ¥^T  =  4<5'3  tons' 
04'O 

If  the  allowable  bearing  pressure  on  the  soil  is  5  tons  per 
square  foot,  the  foundation  should  have  a  width  of 

4-^  =  9-5  ft- 

Thus,  in  Figs.  170  or  172,  the  girders  should  be  about  65  ft. 
long;  extending  approximately  8  ft.  beyond  the  left  column  and 
15  ft.  beyond  the  right  column.  The  beams  in  the  lower  tier 
of  grillage  should  have  a  length  of  about  9.5  ft. 

176.  Case  II.  Projection  at  One  End  Limited  by  Site. — Archi- 
tectural features  frequently  fix  the  length  to  which  the  footing 


FIG.   173- 

may  extend  beyond  one  of  the  end  columns.  Sometimes  it  is 
necessary  to  allow  no  extension  whatever  at  one  end.  When- 
ever either  of  these  two  limitations  arises,  the  footing  may  be  so 
arranged  that  its  pressure  per  foot  of  length  varies  uniformly 
from  one  to  the  other;  as  shown  in  Fig.  173.  This  may  be 
accomplished  by  a  variation  in  the  lengths  of  the  cross-beams, 
resulting  in  a  trapezoidal  area  for  the  footing.  Care  should  be 


278  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VI 

exercised  in  the  choice  and  spacing  of  the  cross-beams  in  order 
that  equal  deflections  at  their  centers  may  be  assured. 

Let  the  intensity  of  the  soil  pressure  per  linear  foot,  at  a  dis- 
tance x  from  the  left  end  be, 

wx  =  Ax  +  B 

so  that  the  soil  pressure  at  the  left  end  will  have  the  intensity  B 
per  linear  foot,  and  at  the  right  end  the  intensity  Al  -\-  B  per 
linear  foot. 

The  static  equations  are 

SF  =  Al2    +  2BI  -  2(P  +  Q  +  R)  =  o  (20) 

SM  =  2Als  +  ^Bl2  -  6aP  -  6(a  +  c)Q 

-  6(a  +  c  +  d)  R  =  o     (21) 

The  elastic  Eq.  (17),  when  developed,  may  be  written  in  the 
form 

c[4(a  +  c)s  +  a{30  +  c)2  +  a($a  +  2c)}]A 
+  d[(a  +  c  +  dY  +  (a  +  c)\2(a  +  c  +  d)2  +  (a  +  c) 


+  5'bO  +  cY  +  a($a  +  2c)]B  +  sd[(a  +  c  +  <2)2  +  (a  +  c) 


=  2o[(c  +  d)2  +  c(c  +  d)]P  +  2od*Q  (22) 

Three  of  the  four  quantities  a,  6,  A  or  B  may  be  determined 
from  these  equations.  A  numerical  value  for  either  a  or  b 
may  be  chosen  arbitrarily  to  suit  the  architectural  features, 
and  the  other  three  quantities  determined  by  a  solution  of 
Eqs.  (20),  (21)  and  (22). 
Illustrative  Problem. 

Let  c  =  12  ft.,  d  —  30  ft.,  P  =  600  tons,  Q  —  goo  tons  and 
R  =  1,500  tons.  The  architectural  features  do  not  allow  a  to 
exceed  6  ft.,  and  the  designer  wishes  to  utilize  this  full  amount. 
In  other  words,  he  fixes  a  =  6  ft.;  and  determines  b,  A  and  B, 
from  the  equations.  Substituting  the  numerical  values  in  (20), 
(21)  and  (22) 

Al2  +  2BI  =  6,000 
2A13  +  sBl2  =  550,800 
8,273,664,4  +  824,040$  =  43,416,000 

whence       I3  —  227.  76i/2  +  i3,884-448/  —  209,928.788  =  o 
or  I  =  22.78  or  66.58  or  138.4  ft. 


SEC.  V 


RESTRAINED   AND    CONTINUOUS   BEAMS 


279 


If  66.58  ft.  is  taken,  then 

A  =  —0.328 
B=    55.98 

and  Al  +  B  =    34.13 

Hence  the  intensity  of  the  soil  pressure  is  55.98  tons  per  linear 
foot  at  the  left  end;  and  decreases  uniformly  to  34.13  tons  at  the 
right  end.  Since  a  =  6  ft.,  and  /  =  66.58  ft.,  the  foundation 
extends  b  =  18.58  ft.  beyond  the  right  column. 

Since  the  quantity  b  does  not  appear  in  Eqs.  (20),  (21)  and 
(22),  it  will  be  found  advantageous  in  any  numerical  problem 
to  arrange  the  nomenclature  so  that  P  may  represent  the  end 
column  beyond  which  the  length  of  the  foundation  is  limited; 
and  assign  a  numerical  value  to  a.  When  no  extension  of  the 
foundation  beyond  P  is  allowed,  a  =  o. 

177.  Case  III.  Both  Projections  Limited  by  Site. — In  the 
case  illustrated  by  Fig.  1 74,  where  both  a  and  b  are  fixed  or  equal 


PIG.  174. 

zero;  the  footing  may  be  so  proportioned  as  to  produce  an 
intensity  of  soil  pressure  per  linear  foot,  which  varies  according 
to  the  ordinates  of  a  parabolic  curve.  The  soil  pressure  per 
square  foot  may  be  made  uniform  by  varying  the  length  of  the 
cross-beams  accordingly.  The  dimensions  a,  b,  c  and  dt  and  the 
loads  P,  Q  and  R  will  be  known;  while  the  three  constants  which 
determine  the  intensity  of  the  soil  pressure  are  to  be  found  by 
solution. 

Let  the  intensity  of  the  soil  pressure  per  linear  foot  at  a 
distance  x  from  the  left  end  be 

wf  =  Ax'2  +  Bx  +  C 


280  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VI 

so  that  the  soil  pressure  at  the  left  end  will  have  the  intensity  C 
per  linear  foot,  and  at  the  right  end  the  intensity  Al2  +  Bl  +  C 
per  linear  foot. 
The  static  equations  are 

SF  =  2AI*  +  3Bl2  +  6CI  -  6(P  +  Q  +  R)  =  o  (23) 

SM  =  3^/4  +  4^3  +  6C72  -  i2aP  -  12(0  +  c)Q  - 

i2(a  +  c  +  d)R  =  o         (24) 

The  elastic  Eq.  (17)  when  developed,  may  be  written  in  the 
form 

c[5(*  +  cY  +  a{4(a  +  c)'  +  a[3(a  +  c)2  +  a(3a  +  2c)}}}A 
+d[(a  +  c  +  dY  +  (a  +  c)  [2(a  +  c  +  d)*  + 

(a  +  c)\s(a  +  c  +  J)2  +  (a  +  c)  {9(a  +  c)  + 
+  c)3  +  a{3(a  +  c)2  +  a(3a  + 
+  c  +  dY  +  (a  +  c)  [i(a  +  c  + 


+  c)  +  2d]\]C 
=  6o[(c  +  d)2  +  c(c  +  d)]P  +  6od*Q  (25) 

Illustrative  Problem. 

Let  a  =  6  ft.,  5  =  12  ft.,  c  =  12  ft.,  d  =  3oft.,P  =  600  tons, 
Q  =  goo  tons  and  R  =  1,500  tons.  Then  I  =  60  ft.  Sub- 
stituting the  numerical  values  in  Eqs.  (23),  (24)  and  (25),  and 
reducing 

i,2oo<4.  +  30$  -f-  C  =  50 
i,8oo,4  +  ^oB  +  C  =  51 
623,028^  +  38,304.6  +  3,8i5C  =  201,000 
whence 

A  =      0.0291 
B  =  -1.646 
C  —    64.466 

and  the  intensity  of  the  soil  pressure  per  linear  foot  at  any 
distance  x  from  the  left  end  is 

wx  =  o.o2gix2  —  1.6462  +  64.466 

The  intensities  of  the  soil  pressure  per  linear  foot  at  the  two 
ends  and  at  five  intermediate  points  are 


SEC.  V 


RESTRAINED   AND    CONTINUOUS  BEAMS 


28l 


X 

w 

X 

w 

Feet 

Tons  per  linear  foot 

Feet 

Tons  per  linear  foot 

o 

64-5 

40 

45-o 

10 

Si-o 

So 

55-o 

20 

43-o 

60 

70.5 

30 

41.0 

If  the  footing  is  to  extend  a  given  distance,  say  5  ft.  beyond 
one  end  column,  and  is  not  to  extend  beyond  the  other  end 
column,  it  will  be  found  advantageous  to  arrange  the  nomencla- 
ture so  that  a  —  o  and  b  =  5.  Equations  (23),  (24)  and 
(25)  are  applicable  when  the  footing  is  not  to  extend  beyond 
either  end  column;  in  which  case  a  =  o  and  6  =  0,  are  to  be 
substituted. 


CHAPTER  VII 
DEFLECTION  OF  TRUSSES 

1  78.  Stress  and  Strain.  —  The  stress  in  any  member  of  a  truss 
under  the  influence  of  external  forces  is  accompanied  by  a  corre- 
sponding strain  or  change  in  length  of  the  member.  The  strains 
which  the  various  members  experience  when  under  stress,  cooper- 
ate in  a  general  distortion  of  the  truss  ;  somewhat  after  the  manner 
illustrated  in  Fig.  175.  The  full  lines  represent  the  configuration 
of  the  truss  when  the  members  are  under  no  strain.  When  loads 
are  applied  gradually  at  £,  C  and  G,  the  truss  undergoes  a  gradual 
distortion  ;  and  finally  conforms  to  the  shape  indicated  by  the 
dotted  lines,  when  the  loads  have  reached  their  full  magnitudes 
FI,  F-z  and  Fs.  The  dotted  outline  gives  an  exaggerated  idea  of 
the  distortion  which  may  be  expected  in  any  practical  problem. 

The  first  step  in  finding  the  movement  of  displacement  of  any 
point  in  the  truss,  is  the  determination  of  the  strain  in  each 
member  subjected  to  stress.  This  is  easily  accomplished  by 
the  well  established  law  of  mechanics,  viz.: 

unit  stress  ,  ,      ,  ~  .     ,%     ,    ,     ..  ., 

-  =  modulus  (measure  or  coefficient)  of  elasticity. 
unit  strain 

Let  P  =  the  stress  in  a  member  in  pounds. 

A  =  the  area  of  cross-section  of  the  member  in  square 

inches  (inch2). 

/  =  the  length  of  the  member  in  inches, 
D  =  strain,  or  change  in  length,  of  the  member  in  inches. 

d  =~Y  =  unit  strain  in  inches  per  inch  (inch/inch),  a  ratio. 

p 

j-  =  unit  stress  in  pounds  per  square  inch  (pounds/inches2). 

A 

E  =  modulus    of    elasticity    in    pounds    per    square   inch 

(pounds/inches.2) 
P 
A  PI 


7 

282 


SEC.  I 


DEFLECTION    OF    TRUSSES 


283 


Hence  the  strain  or  change  in  length  of  a  member  may  be 
determined  if  its  length  and  cross-sectional  area,  the  stress 
which  it  resists  and  the  modulus  of  elasticity  of  its  material 
are  known.  Experiments  show  that,  for  any  given  material, 
the  modulus  of  elasticity  is  approximately  constant  for  all 
unit  stresses  below  a  certain  limit,  called  the  elastic  limit. 
The  elastic  limit  for  structural  steel  is  about  60  per  cent  of  its 
ultimate  strength;  hence  the  permissible  unit  stress  in  all 
current  practice  is  well  within  the  elastic  limit.  The  modulus 
of  elasticity  for  ordinary  structural  steel  is  about  29,000,000 
Ib.  per  square  inch;  hence  if  a  member  50  ft.  long,  having  a 
cross-sectional  area  of  20  sq.  in.,  is  subjected  to  a  tensile  stress 
of  300,000  Ib.,  the  strain  or  elongation  of  the  member  will  be 

~       300,000  X  50  X  12 

D  =  * — '-—-  -  =  0.31  in. 

20  X    29,OOO,OOO 

SEC.  I.     ALGEBRAIC  METHOD 

179.  The  algebraic  solution  may  be  developed  by  equating 
the  external  work  done  by  the  external  forces,  to  the  internal 


work  performed  upon  the  members  of  the  truss. 

Let  AI  (Fig.  175)  represent  the  vertical  component  of  the  dis- 
placement, or  deflection  of  the  point  B  in  the  direction  of  the  force 
FI.  During  the  movement  of  the  point  from  B  to  B',  the  force 
which  has  been  increased  from  zero  to  FI  performs  work  to  the 
amount  of  %  FiAi.  Each  member  under  stress,  and  thereby 
subject  to  strain,  contributes  its  share  to  the  deflection  of  B. 
Consider  any  member  HK,  for  example,  which  is  /  inches  long 
and  has  a  cross-sectional  area  of  A  sq.  in.  Let  PI,  P2  and  P3 


284  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VII 

represent  the  stresses  in  HK,  caused  by  the  loads  FI,  F2  and 
Fz  respectively;  and  let  PI  -f  P2  +  PZ  =  P  =  the  total  stress. 

PI 
The  strain  in  HK  is  D  =  ^ 

and  the  total  work  performed  upon  HK,  as  the  three  loads  are 
applied  and  the  stresses  gradually  increased  from  zero  to  their 
final  values,  is 


2V  2  2        AE 

The  work  performed  upon  HK,  resulting  from  the  load  FI  is 

Ipn=  JP   JW 

2  2    l  AE 

PI 
Let  2  %  Pi-rjf  represent  the  sum  of  the  work  performed  upon 


each  member  by  the  stress  resulting  from  the  load  FI.  Since 
the  external  work  done  by  the  loads  must  equal  the  internal 
work  performed  upon  the  members,  then 

IF   A     _  2ip    PI 
-  r\  AI  —  L-  r\  —  -=, 

2  2          AE 

Whence  PI    Pl_ 

F!  .4£ 

in  which  PI  is  the  stress  in  any  member  due  to  the  load  FI,  and 
P  is  the  stress  in  the  same  member  due  to  the  three  loads  FI,  F% 
and  FZ. 

Since  the  stress  PI  in  any  member  varies  directly  as  the  load 

p 

Fi;  it  is  obvious  that  the  ratio  ^r  is  constant  for  that  member 

•TI 

for  all  values  of  FI. 


To  get  the  value  of  the  ratio  u  for  each  member,  assume  for 
the  present  that  FI  equals  i  lb.;  in  other  words,  place  i  lb.  at 
the  point,  acting  in  the  direction  of  the  desired  deflection  and 
compute  the  stresses  in  all  members  for  this  i  lb.  loading. 
The  resulting  stresses  expressed  in  pounds,  when  divided  by 
i  lb.,  will  represent  the  ratio  u  for  the  various  members.  For 


SEC.  I 


DEFLECTION    OF    TRUSSES 


any  member,  the  stress  P  and  the  ratio  u  have  positive  or 
negative  signs  corresponding  to  tension  or  compression.  The 
strain  in  any  member  causes  the  point  in  question  to  deflect 
in  the  direction  in  which  the  i  Ib.  load  is  assumed  to  act,  if  P 
and  u  have  like  signs ;  and  in  the  opposite  direction,  if  they  have 
unlike  signs. 

In  Eq.  (i)  E  is  the  only  quantity  which  is  constant  for  all 
members  of  the  truss,  provided  all  members  are  made  of  the 
same  material  (which  is  usually  the  case).  Hence  the  expres- 

Pul 
sion  — r-  may  be  tabulated  and  computed  for  each  member; 

Pul 

and  the  deflection  found  by  dividing  the  algebraic  sum  I  — T- 

A. 

by  E,  as  in  the  following  problem: 
1 80.  Illustrative  Problem. 

The  truss  in  Fig.  176  supports  a  load  of  240,000  Ib.  at  each 
bottom  chord  panel  point.  In  Table  I,  the  length  of  each  mem- 


FIG.  176. 

ber  is  given  in  column  i ;  the  gross  cross-sectional,  in  column  2 ; 
and  the  stress,  in  column  3.  The  quantities  in  column  4,  when 
divided  by  E,  represent  the  strains.  To  find  the  deflection  at 
point  6,  place  i  Ib.  at  6  and  compute  the  stresses;  as  given  in 
Fig.  177.  These  stresses,  when  divided  by  i  Ib.,  are  the  ratios 
UQ  in  column  5.  In  this  case  P  and  MQ  have  like  signs  for  each 
member,  hence  the  quantities  in  column  6  are  all  positive,  and 
their  algebraic  sum  is  42,246.  Since  P  was  expressed  in  units 
of  1,000  Ib.,  the  deflection  at  point  6  is 

42,246  X  1,000       42,246,000  .,  . 

A6  =  —      — =  -  -  =  1.46  in. 

E  29,000,000 


286 


THEORY    OF    FRAMED    STRUCTURE 


CHAP.  VII 


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SEC.  I 


DEFLECTION    OF    TRUSSES 


287 


In  like  manner  the  deflection  at  point  4  is  found  by  placing  a 
load  of  i  Ib.  at  4  (Fig.  178)  and  obtaining  the  ratios  u±  in 
column  7.  The  products  of  corresponding  values  in  columns 
4  and  7  are  given  in  column  8.  In  this  instance,  P  and  u±  have 
unlike  signs  for  the  member  4-5;  hence  the  quantity  represent- 
ing the  member  4-5  in  column  8  has  the  negative  sign,  and  the 
strain  in  that  member  tends  to  raise  the  point  4.  The  alge- 

1       -3/4       3     -%         5     -%       7     -3/4         9 


I 


8       •/-. 


<o 6)  27 '=162' 
FIG.  177 


J 


3       -/ 


FIG.  118 


braic  sum  of  the  quantities  in  column  8  is  31,916;  and  the 
deflection  of  point  4  is 


Similarly  the  deflection  of  point  6  is 

22,68^,000 

A6  =  — '- — ^    -  =  0.78  in. 
29,000,000 

Since  the  design  and  loading  of  the  truss  in  this  problem  are 
both  symmetrical  about  the  center  line,  it  is  obvious  that  the 
vertical  deflections  at  the  points  8  and  10  are  respectively  the 
same  as  at  points  4  and  2. 

The  horizontal  displacement  of  any  point  may  be  obtained 


288 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  VII 


in  a  similar  manner.  Suppose  that  the  horizontal  displacement 
of  the  point  i  in  Fig.  176  is  required,  when  the  truss  is  held  fast 
at  the  left  support  and  rests  on  rollers  at  the  right  support;  as 
shown  in  Fig.  175.  The  loads  are  assumed,  as  in  Fig.  176; 

PI 

and  the  values  of  -j-  of  column  4,  Table  I,  are  therefore  appli- 
A 

cable.  Place  i  Ib.  at  point  i,  acting  horizontally  either  to  the 
right  or  left,  let  us  say  to  the  right  (as  in  Fig.  179);  compute 


10 


0.222 i 


FIG.  179. 


a???1 


the  reactions   and   tabulate  the  resulting  ratios  in  a  column 

marked  u\\  and  make  the  extensions  -  j— • 

A. 

The  horizontal  displacement  of  the  point  i  will  be 

".V  Al  =  2S 

Since  the  i  Ib.  load  was  taken  as  acting  to  the  right,  a  positive 
value  for  the  summation  will  indicate  a  displacement  to  the 
right;  and  vice  versa. 

SEC.  II.     GRAPHIC  METHOD 

181.  Williot  Diagrams. — The  graphic  method  of  finding  the 
deflections  of  a  truss  is  accomplished  by  drawing  a  Williot 
diagram ;  after  the  strain  in  each  member  has  been  determined, 
as  described  in  Sec.  I.  In  order  to  compare  the  results  of  the 
algebraic  method  with  the  graphic,  the  latter  will  be  explained 
in  connection  with  the  problem  illustrated  in  Fig.  176.  Since 
the  modulus  of  elasticity  is  constant  for  all  members,  the 
quantities  in  column  4  of  Table  I  will  be  taken  to  represent  the 
strains;  and  the  factors  1,000  and  E  =  29,000,000  Ib.  per 


SEC. 


DEFLECTION    OF    TRUSSES 


289 


square  inch  will  be  introduced  at  the  end  of  the  problem,  as  in 
the  algebraic  solution.  Three  solutions  will  be  given,  based 
on  three  different  assumptions. 

First  Solution. — We  shall  assume  that  the  point  6  is  fixed  in 
position  and  the  member  5-6  fixed  in  direction ;  and  determine 
the  relative  displacements  of  the  other  points,  with  reference 
to  point  6,  when  the  various  members  are  subject  to  the  strains, 
as  indicated  in  column  4,  Table  I.  First  let  us  consider  the 
triangular  unit  4-5-6  of  the  truss  in  Fig.  1 76.  For  convenience 
this  unit  is  shown  in  Fig.  180,  and  designated  as  abc.  From 
column  4  the  strains  are  represented  by  the  following  quanti- 


ties, be  =  +4,610;  ac  =  +3,150  and  ab  =  —2,680.  According 
to  the  hypothesis,  the  point  c  is  to  remain  fixed  in  position  and 
the  member  be  is  to  remain  fixed  in  direction. 

The  strain  in  be  is  an  increase  in  length,  represented  by  the 
quantity  4,610;  consequently  from  b  we  lay  off  to  any  conveni- 
ent scale  (not  necessarily  the  scale  used  in  laying  off  be)  bd  = 
4,610,  above  the  point  b.  The  point  which  was  originally  at 
b  has  now  moved  to  d.  The  location  of  d  was  a  simple  matter; 
since  according  to  the  hypothesis,  the  member  be  (when  elongat- 
ing) had  no  option  except  to  extend  vertically  upward. 

The  location  of  the  new  position  of  the  point  at  a  is  more  com- 
plicated. The  member  ac  lengthens;  and  since  c  is  fixed,  the 
extension  is  to  the  left  of  and  away  from  c.  Hence  we  lay  off  ae 
=  3,150.  The  movement  of  a  is  also  influenced  by  the  member 
ab.  Hence  of  is  laid  off  equal  and  parallel  to  bd,  and  fd  is 


2QO 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VII 


drawn  to  represent  the  original  length  of  ab.  The  member  ab 
is  shortened  by  the  strain  2,680;  since  the  point  d  has  been 
located;  therefore,  the  point /moves  towards  d,  the  amount /g 


0    +3080      I    +3080      4     +3/50       6"    +3150       8 


=  2,680.  Since  the  two  members  ac  and  ab  are  connected;  the 
two  points  e  and  g,  which  represent  the  ends  of  these  members, 
when  under  strain,  must  coincide.  Therefore  an  arc  having 
the  radius  ce  is  described  about  c  as  a  center,  and  an  arc  having 


SEC.  II  DEFLECTION  OF  TRUSSES  2pl 

the  radius  dg  is  described  about  d  as  a  center.  The  intersection 
of  the  two  arcs  at  h  marks  the  new  position  of  the  point,  which 
was  originally  at  a.  Since,  in  any  practical  problem,  the  arcs 
are  very  small  in  proportion  to  their  radii;  it  will  be  sufficiently 
accurate  to  draw  the  straight  lines  eh  and  gh  perpendicular  to 
ac  and  ab,  respectively. 

Reference  to  Fig.  181  shows  at  once  that  the  displacement  of 
the  points  a  and  b  may  be  determined,  without  including  in  the 
diagram  the  members  themselves.  Let  the  point  c,  which  is 
assumed  fixed  in  position  be  the  origin  or  fixed  point  of  refer- 
ence. The  member  cb  lengthens,  and  the  point  b  moves  upward 
and  away  from  c;  therefore  lay  off  cb  =  4,610,  upward  from  c. 
The  member  ca  also  lengthens,  and  the  point  a  moves  hori- 
zontally to  the  left  from  c\  therefore  lay  off  ca'  =  3,150,  to  the 
left  from  c.  The  member  ab  shortens,  and  the  point  a  moves 
(also)  diagonally  upward  and  to  the  right  towards  5;  therefore 
lay  off  ba"  =  2,680,  parallel  to  ab  upward  and  to  the  right 
from  b.  The  perpendiculars  through  a'  and  a"  intersect  at  a. 
Since  c  is  the  origin  or  reference  point,  the  movements  or 
displacements  of  the  points  a  and  b  from  their  original  position 
are  indicated  by  their  positions  relative  to  the  origin  c. 

The  procedure  is  similar  when  the  frame  consists  of  a  series 
of  triangular  units,  as  illustrated  by  the  truss  in  Fig.  182.  It 
is  assumed  that  the  truss  and  loads  are  the  same  as  shown  in 
Fig.  176,  and  specified  in  Table  I.  The  quantities  which 
represent  the  strains  in  the  various  members,  taken  from 
column  4  of  the  table,  are  indicated  in  the  figure  for  convenience. 
A  positive  sign  indicates  elongation,  and  a  negative  sign  indi- 
cates a  shortening.  Each  strain  is  laid  off  in  the  Williot  dia- 
gram (Fig.  183)  parallel  to  the  original  direction  of  the  member 
in  which  the  strain  occurs.  Joint  6  is  assumed  fixed  in  position, 
and  the  member  5-6  is  assumed  fixed  in  direction;  hence  the 
point  6  (Fig.  183)  is  the  origin  or  reference  point  in  the  Williot 
diagram. 

Joint  5  moves  upward  from  joint  6,  hence  point  5  is  laid  off  to 
a  convenient  scale  abovq  point  6.  Joint  4  moves  to  the  left 
away  from  joint  6,  and  upward  to  the  right  toward  joint  5; 
hence  the  strain  in  4-6  is  laid  off  to  the  left  from  point  6,  and  the 


THEORY   OF   FRAMED    STRUCTURES  CHAP.  VII 

strain  in  4-5  is  laid  off  upward  and  to  the  right  from  point  5. 
The  perpendicular  lines  drawn  from  the  extremities  of  these  two 
strains  intersect  at  point  4.  The  position  of  point  .4,  relative 
to  the  reference  point  6,  indicates  the  movement  of  point  4  from 
its  original  position. 

Joint  3  moves  to  the  right  toward  joint  5,  and  moves  neither 
upward  nor  downward  with  reference  to  joint  4,  since  there  is  no 
strain  in  the  member  3-4;  hence  the  strain  in  3-5  is  laid  off  to  the 
right  from  point  5,  and  no  strain  is  laid  off  from  point  4.  The 
perpendicular  lines  drawn  from  the  extremity  of  the  strain  in 
3-5,  and  from  point  4,  intersect  at  point  3.  The  position  of 
point  3,  relative  to  point  6,  indicates  the  movement  of  joint  3 
from  its  original  position. 

The  remaining  joints  i,  2  and  o  are  similarly  treated  in 
consecutive  order;  and  the  joint  o  is  finally  located  in  the 
diagram. 

The  scale  used  in  constructing  the  Williot  diagram  is  too  large 
to  be  of  service  in  locating  the  movement  or  displacement  of 
each  joint  with  reference  to  the  truss  itself.  If  a  smaller  and 
more  convenient  scale  is  taken,  the  displacements  may  be 
indicated  by  the  dotted  outline  as  shown  in  Fig.  182;  although, 
even  there,  the  actual  distortion  is  greatly  exaggerated  in 
comparison  to  the  dimensions  of  the  truss.  On  account  of 
symmetry  in  loading  and  design,  the  distortion  of  the  whole 
truss  may  be  determined  by  drawing  the  Williot  diagram  for 
either  half.  The  dotted  outline  gives  a  true  conception  of  the 
distortion,  when  joint  6  is  fixed  in  position,  the  member  5-6 
fixed  in  direction,  and  the  members  are  subjected  to  the  strains 
as  indicated.  It  is  evident,  however,  that  joints  o  and  12, 
at  the  points  of  support  instead  of  joint  6,  remain  fixed  in 
elevation.  The  correction  for  this  error  in  assumption  is  made 
by  moving  the  dotted  outline  vertically  downward  until  the 
joints  a  and  b  are  at  the  same  elevation  as  o  and  12.  The 
corresponding  correction  is  made  in  the  Williot  diagram  by 
considering  A  as  the  origin,  or  reference  point,  instead  of  point 
6.  The  vertical  displacements  or  deflections  of  joints  2,  4  and 
6  are  obtained  by  scaling  dz,  d±  and  J6;  multiplying  each  by 
1,000  and  dividing  by  29,000,000;  as  in  the  algebraic  solution. 


SEC.  II 


DEFLECTION    OF   TRUSSES 


293 


The   horizontal   displacement   of   any   point   is   obtained   by 
scaling  its  horizontal  distance  from  the  line  A6.     Thus  the 


!       -3380 


FIG.  185 


horizontal,  as  well  as  vertical  displacements  of  all  joints  in  the 
truss,  may  be  found  by  drawing  one  Williot  diagram. 

Second  Solution. — If  the  loading  or  design  in  the  first  solution 


2Q4  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VII 

had  not  been  symmetrical,  the  assumption  that  the  member 
5-6  remained  fixed  in  direction  would  not  have  been  true;  and  a 
correction  would  have  been  necessary.  The  method  by  which 
this  correction  may  be  made  will  be  illustrated  by  using  the  same 
strains  as  before,  and  assuming  that  the  joint  o  (Fig.  184) 
remains  fixed  in  position,  and  the  member  o-i  fixed  in  direction. 
The  Williot  diagram  is  shown  in  Fig.  185,  where  the  point  o  is 
the  origin  or  reference  point.  The  point  i  is  located  first;  then 
2,  4,  3,  5  and  so  on  to  point  12. 

The  distorted  outline  may  now  be  drawn,  as  shown  by  the 
dotted  lines  in  Fig.  184.  This  outline  has  precisely  the  same 
configuration  as  the  distorted  outline  of  Fig.  182.  In  order  that 
the  distorted  outline  may  truly  represent  the  deflections  of  the 
various  joints,  it  is  necessary  that  the  dotted  figure  be  rotated 
about  joint  o;  until  the  joint  12  of  the  dotted  outline  is  on  a 
level  with  joint  1 2  of  the  original  outline.  During  this  rotation, 
the  vertical  movement  of  joint  12  of  the  dotted  outline  is 
represented  by  the  vertical  distance  from  point  12  to  point  o 
in  the  Williot  diagram.  Let  r  represent  this  vertical  distance. 
Draw  the  horizontal  line  12-12',  intersecting  the  vertical  line 
0-12'  at  12'.  Then  r  is  the  distance  0-12'.  Taking  0-12'  as  the 
bottom  chord,  draw  the  truss  to  scale  as  shown  by  dotted  lines 
in  Fig.  185.  This  dotted  outline  is  called  the  Mohr  rotation 
diagram,  which  supplements  the  Williot  distortion  diagram. 
The  total  vertical  displacement  or  deflection  of  any  joint  may  be 
considered  as  the  result  of  two  operations — distortion  and 
rotation.  As  the  distortion  takes  place,  joint  2  is  lowered  the 
vertical  distance  from  point  o  to  point  2  in  the  distortion 
diagram.  As  the  distorted  outline  is  rotated  about  joint  o, 
and  joint  12  moves  through  the  vertical  distance  r,  joint  2  is 
lowered  the  additional  distance  }/§r,  represented  by  the  distance 
2r-o  in  the  rotation  diagram.  Hence  the  total  deflection  of 
joint  2  is  represented  by  the  distance  di,  which  scales  22,700. 

Joint  10  is  raised  by  distortion  the  vertical  distance  from  point 
o  to  point  10 ;  and  lowered  by  rotation  y§r  or  the  distance 
lo'-o.  Hence  the  net  deflection  of  joint  10  is  represented  by  the 
distance  di0,  which  also  scales  22,700.  Similarly  the  vertical 
displacement  of  joint  8  is  represented  by  the  vertical  distance 


SEC.  II  DEFLECTION    OF   TRUSSES  295 

from  point  8'  in  the  rotation  diagram,  to  point  8  in  the  distortion 
diagram;  and  the  horizontal  displacement  of  joint  9  is  the 
horizontal  distance  from  point  9'  in  the  rotation  diagram,  to 
point  9  in  the  distortion  diagram. 

It  will  be  seen  by  inspection  that  the  vertical  displacements  of 
all  joints  as  obtained  from  Fig.  185  are  the  same  as  given  in 
Fig.  183,  or  by  the  algebraic  solution  in  Table  I.  The  hori- 
zontal displacements  differ,  because  in  Fig.  183  the  truss  was 
held  at  joint  6  instead  of  at  joint  o,  as  in  Fig.  185. 

Third  Solution. — In  the  first  solution,  a  translation  of  the 
distorted  outline  (Fig.  182)  was  necessary,  because  of  the 
erroneous  assumption  of  a  joint  fixed  in  position;  while  in  the 
second  solution,  a  rotation  of  the  distorted  outline  (Fig.  184) 
was  necessary,  because  of  the  erroneous  assumption  that  a 
member  was  fixed  in  direction.  The  Williot  distortion  dia- 
gram, in  Fig.  187,  is  drawn  on  the  assumption  that  the  joint  4 
is  fixed  in  position  and  the  member  4-1  is  fixed  in  direction. 
Point  4  is  the  origin;  and  the  other  points  may  be  located  in  the 
order  i,  2,  o,  3,  5,  6  and  so  on  to  12.  The  distorted  outline  of 
the  truss,  shown  in  Fig.  186,  has  precisely  the  same  configura- 
tion as  in  Figs.  182  and  184.  The  Mohr  rotation  diagram  will 
be  drawn  on  the  basis  of  a  fixed  support  at  o,  and  a  roller  support 
at  12.  It  is  apparent  that  the  distorted  outline  must  be  trans- 
lated vertically  downward,  then  horizontally  to  the  right; 
until  joint  o  of  the  distorted  outline  coincides  with  joint  o 
of  the  original  outline.  The  distorted  outline  must  then  be 
rotated  about  joint  o,  until  joint  12  of  the  distorted  outline  is 
level  with  joint  1 2  of  the  original  outline.  Since  joint  o  is  now 
to  be  considered  fixed  in  position,  instead  of  joint  4,  the  origin 
is  moved  from  point  4  to  point  o  in  Fig.  187.  Draw  the  vertical 
through  point  o,  and  the  horizontal  through  point  12,  intersect- 
ing at  12';  and  on  0-12',  as  the  bottom  chord,  construct  the 
truss  to  scale. 

The  total  vertical  displacement  or  deflection  of  any  joint  may 
be  considered  as  the  result  of  three  operations — distortion, 
vertical  translation  and  rotation.  As  the  distortion  takes  place, 
joint  2  is  lowered  the  vertical  distance  from  point  4  to  point  2 
in  the  distortion  diagram.  The  vertical  translation  lowers  the 


296 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  VII 


joint  the  vertical  distance  from  point  o  to  point  4.     Finally  the 
rotation  lowers  the  joint  the  vertical  distance  from  point  2'  to 


+3080 


4     +3150        6    +3150       8     +3080      10   +3080 
FIG.  186 


FIG.  1 87 


point  o.  Hence,  the  total  vertical  distance  through  which  the 
joint  2  moves,  is  represented  by  the  vertical  distance  from  point 
2r  in  the  translation  rotation  diagram  to  point  2  in  the  distortion 
diagram;  which  is  d2  =  22,700.  The  vertical  displacements  or 


SEC.  Ill 


DEFLECTION   OF   TRUSSES 


2Q7 


deflections  here  found  agree  with  those  of  the  previous  solu- 
tions. The  horizontal  displacements  agree  only  with  those  of 
the  second  solution,  where  joint  o  was  also  subject  to  horizontal 
restraint 

SEC.  III.     GENERAL  CONSIDERATIONS 

Deflections  are  the  result  of  changes  in  length  of  the  members 
or  distances  between  panel  points;  whether  caused  by  strains, 
temperature,  or  play  between  the  pins  and  pin-holes.  The 
treatment  in  all  cases  is  the  same  after  the  changes  in  length 
have  been  determined. 

182.  Temperature. — Suppose  that  we  wish  to  determine 
vertical  displacement  of  the  point  4,  Fig.  176;  when  the  tempera- 
ture of  the  top  chord  is  10  degrees  above  normal  and  the  tem- 
perature of  the  bottom  chord  is  15  degrees  below  normal.  The 
coefficient  of  thermal  expansion  is  0.0000065.  Consider  the 
member  1-3.  Its  change  in  length,  due  to  10  degrees  rise  in 
temperature,  is 

324  X  10  X  0.0000065  =  +0.021  in. 

This  change  in  length  is  treated  precisely  as  if  it  were  a  strain 
resulting  from  the  stress  in  the  member,  due  to  loads.  Since 
the  coefficient  of  thermal  expansion  is  the  same  for  all  members, 

TABLE  II 


Length  in 

Temperature 

Member 

inches 

change 

It 

u\ 

ItUi 

I 

t 

i-3 

324 

+  10 

+3,240 

—  i.o 

-3,240 

3-5 

324 

+  10 

+3  ,  240 

—  1  .0 

-3,240 

5-7 

324 

+  10 

+3,240 

-0.5 

—  1,620 

7-9 

324 

+  10 

+3  ,  240 

-0.5 

—  i  ,620 

0-2 

324 

-i5 

-4,860 

+0.5 

-2,430 

2-4 

324 

-15 

-4,860 

+0.5 

—  2,430 

4-6 

324 

-i5 

-4,860 

+0-75 

-3,645 

6-8 

324 

-IS 

-4,860 

+0-75 

-3,645 

8-10 

324 

-15 

-4,860 

+0.25 

-1,215 

IO-I2 

324 

-IS 

-4,860 

+0.25 

-1,215 

—24,300 


298 


THEORY   OF   FRAMED    STRUCTURE 


CHAP.  VII 


the  computation  may  be  arranged  as  in  Table  II.     The  vertical 
displacement  of  point  4  is 

A4  =  2//W4  X  0.0000065  =  —o.i  6  in. 

The  negative  sign  indicates  that  the  point  4  is  raised  by  the 
effect  of  the  temperature. 

The  problem  may  be  solved  graphically  by  drawing  a  Williot 
diagram  and  using  the  quantities  It  in  Table  II,  to  represent  the 
changes  in  length. 

183.  Camber. — It  is  desirable  that  the  truss  in  Fig.  175,  when 
loaded,  shall  have  the  configuration  represented  by  the  full 
lines,  rather  than  by  the  dotted  lines.  In  order  to  accomplish 


PIG.   188. 

this  end,  it  is  necessary  to  camber  the  truss,  by  increasing  the 
length  of  each  compression  member  and  decreasing  the  length 
of  each  tension  member  by  the  amount  of  strain  which  it 
experiences  under  load.  Thus,  in  Fig.  176,  the  strain  in  the 
member  1-3  from  Table  I  is 

-3,380  X  1,000  =_oii6.n 

29,000,000 

This  member  is  shortened  about  one-eighth  of  an  inch 
when  the  truss  is  loaded,  hence  its  original  length  is  made  27'- 
oj£".  If  all  members  are  treated  accordingly,  and  the  truss 
erected  on  false  work,  in  such  a  way  that  the  members  are 
carrying  little  or  no  stress;  the  truss  will  have  the  configuration 
of  Fig.  188;  which  is  known  as  a  camber  diagram.  The  camber 
diagram  is  constructed  from  a  Williot  diagram,  drawn  by 
using  the  strains  as  given  in  column  4  of  Table  I,  with  opposite 
signs.  Trusses  are  usually  cambered  for  dead  load  plus  live 
load,  impact  not  included;  sometimes  the  dead  load  plus  two- 
thirds  the  live  load  is  taken. 


SEC.  Ill 


DEFLECTION    OF    TRUSSES 


299 


The  following  approximate  method  is  sometimes  used.  If 
the  average  unit  stress  in  the  members  is  14,000  Ib.  per  square 
inch,  based  on  the  gross  section;  the  strain  in  every  10  ft.  of 
length  is  a  little  short  of  one-sixteenth  of  an  inch.  Hence  a 
truss  may  be  cambered  by  a  rule-of-thumb  method,  if  one- 
eighth  of  an  inch  for  every  10  ft.  in  length  is  added  to  the  top 
chord;  the  length  of  all  other  members  remaining  the  same  as 
if  no  allowance  were  made.  Thus  each  top  chord  member  of 
the  truss  in  Fig.  176  would  be  made  27/-oJHV/  long- 

184.  Maxwell's  Theorem  of  Reciprocal  Deflection. — The 
proof  of  Maxwell's  theorem  as  applied  to  beams,  on  page  231,  is 

C  Pa     D 


^*e^ J&~^ 

-- -^.- =.--.-<**— 68)  ?0'=1?0'— ->| 


FIG.  189 


C  Vg      D 


F/G.A92 


equally  applicable  to  trusses.  It  is  also  easily  proven  by  the 
algebraic  method  of  Sec.  I.  Let  A0  represent  the  deflection  at 
A ,  caused  by  the  load  F&tB  (Fig.  189) ;  and  let  A6  represent  the 
deflection  at  B,  caused  by  the  load  F  at  A  (Fig.  191);  then 

PaUal 


Aa= 


AE 


and 


AE 


300  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VII 

where  Pa  =  stress  in  any  member  of  Fig.  189. 

Pb  =  stress  in  any  member  of  Fig.  190. 

ua  =  stress  in  any  member  of  Fig.  191. 

ub  =  stress  in  any  member  of  Fig.  192. 
Choose  any  member  as  CD,  then 

D         2F  SF  8       ,  2 

Pa  =  — ,  Pb  =  — ,  ua  =  -  and  ub  =  - 
999  9 

hence       Paua  =  Pbub 

If  any  other  member  be  taken  at  random  it  will  be  found  that 

PaUa   =    PbUb 

therefore  Aa  =  A6 

or  the  deflection  at  A,  caused  by  a  load  at  B,  equals  the  deflec- 
tion at  B  caused  by  the  same  load  at  A . 


CHAPTER  VIII 
SWING  BRIDGES 

A  swing  bridge  rotates  in  a  horizontal  plane  about  a  vertical 
axis,  and  is  classified  as  center-bearing  or  rim-bearing  in  accord- 
ance with  the  method  by  which  it  is  supported  while  swinging. 

SEC.  I.     CENTER-BEARING  SWING  BRIDGES 

185.  General  Considerations. — When  a  bridge  of  this  type 
is  open,  each  truss  is  supported  at  the  end  of  a  cross-girder 
which  rests  on  a  center-bearing  c  (Fig.  193).     This  bearing  is 
usually  a  phosphor-bronze  disk,  from  i  to  3  ft.  in  diameter, 
between  two  hardened  steel  disks.     To  prevent  the  bridge  from 
tipping;  balance  wheels  w,  about  18  in.  in  diameter  and  six  to 
eight  in  number,  are  fastened  to  the  trusses  and  floor  system  in 
such  a  way  that  they  roll  on  a  circular  track  /.     These  wheels 
are  so  adjusted  that  they  carry  no  load,  except  when  the  bridge 
is  out  of  balance  on  account  of  wind  pressure  or  similar  causes. 

When  the  bridge  is  closed;  the  ends  a,  b}  d  and  e  are  raised  a 
proper  amount  by  wedges.  Wedges  are  also  inserted  at  / 
and  g  a  sufficient  amount  to  give  the  trusses  a  bearing  on  the 
pier  independent  of  the  cross-girder;  but  no  attempt  is  made  to 
lift  the  trusses  at  these  points  in  order  to  relieve  the  cross- 
girder  of  any  of  the  dead  load.  Thus  at  the  center  support, 
the  dead  load  is  carried  by  the  cross-girder ;  while  the  live  load 
is  supported  directly  by  the  pier. 

1 86.  Conditions  of  Loading. — When  the  bridge  is  open,  the 
dead  load  is  balanced  about  the  center  support.     When  the 
bridge  is  closed,  the  total  dead  load  reaction  at  the  center  is  re- 
lieved, as  the  wedges  are  driven  and  the  ends  raised.     (If  the 
ends  were  raised  to  a  sufficient  height,  the  bridge  would  be  lifted 
free  of  the  center  support;  and  the  reaction  which  formerly 
supported  the  bridge  would  be  transferred  from  the  center 

301 


302 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  VIII 


to  the  ends.)  The  mechanical  parts  of  the  bridge  are  usually 
designed  in  such  a  way  that  the  end  wedges,  when  fully  driven, 
will  provide  a  positive  dead-load  reaction;  somewhat  greater 
than  the  negative  live-load  and  impact  reaction.  This  insures 
the  bridge  against  hammering  of  one  end  when  the  train 
comes  on  at  the  other  end.  By  this  arrangement,  the  dead- 
load  end  reactions  are  less  than  those  obtained  theoretically 
when  the  truss  is  considered  as  a  continuous  beam  over  three 
level  supports.  The  cost  of  installation  and  operation  are 
thereby  reduced. 

If  hammering  is  eliminated  and  the  ends  remain  fixed  in 


x/«  

1  1 

~~l»jy  — 

fW                ,                                   .r. 

FIG.  1 93 


FIG.  194 


FIG.  195 


FIG.  136 

elevation,  the  live-load  reactions  may  be  computed  on  the  basis 
of  complete  continuity  of  each  truss.  If,  however,  on  account 
of  error  in  design  or  adjustment  of  the  wedges;  or  because  of 
settlement  of  the  piers,  it  happens  that  no  dead-load  end 
reaction  is  present,  we  have  the  condition  as  illustrated  in 
Fig.  194.  Let  us  assume,  for  example,  that  a  very  small 
clearance  exists  between  the  beam  and  its  end  supports.  The 
total  weight  of  the  beam  is  supported  at  C,  as  is  the  case  when 
the  bridge  is  open.  When  the  live  load  comes  on  the  arm  BC 
(Fig.  195),  the  beam  tilts  until  it  finds  a  bearing  at  B\  and  the 
live  load  is  supported  by  BC  acting  as  a  simple  span.  However, 
the  dead  load  of  weight  of  the  beam  is  still  entirely  supported 


SEC.  I  SWING  BRIDGES  303 

at  C.  If  the  live  load  is  present  in  both  arms,  (Fig.  196)  the 
beam  will  deflect  until  it  has  a  bearing  at  A  and  B ;  and  thus  a 
condition  of  continuity,  or  partial  continuity,  must  be  considered 
in  finding  the  live  load  reactions.  The  extent  of  the  continuity 
will  depend  upon  the  amount  of  clearance  which  previously 
existed  at  A  and  B  in  Fig.  194.  The  dead  load  is  still  totally 
supported  at  C.  In  order  to  provide  for  these  contingencies,  the 
following  conditions  or  classes  of  loading  should  be  considered. 

Case  I. — Dead  load;  bridge  open  or  wholly  supported  at  the 
center. 

Case  II. — Dead  load;  bridge  closed,  both  ends  raised  until  a 
definite  end  reaction  is  attained;  amount  to  be  specified  after 
negative  live  load  and  impact  end  reaction  have  been  deter- 
mined. ' 

Case  III. — Live  load  on  one  arm  only;  acting  as  a  simple  span. 

Case  IV. — Live  load  on  one  arm  only;  continuous  girder 
action. 

Case  V. — Live  load  on  both  arms;  continuous  girder  action. 

If  the  ends  are  raised  we  have  Case  II,  combined  with  Case  IV 
or  Case  V.  If  the  ends  are  not  raised  we  have  Case  I,  com- 
bined with  Case  III  or  Case  V. 

The  bottom  chords  are  usually  better  protected  from  the 
rays  of  the  sun  than  the  top  chords.  Because  of  this,  and  also 
on  account  of  cold  weather  and  ice,  it  is  apparent  that  the  tem- 
perature of  the  top  chord  may  be  considerably  higher  than  that 
of  the  bottom  chord.  Thus  the  top  chord  may  be  lengthened 
and  the  bottom  chord  shortened,  causing  the  span  to  "hump" 
at  the  center;  thereby  relieving  the  center  reactions  somewhat 
and  increasing  those  at  the  ends. 
Except  in  special  cases  the  temperature  factor  is  neglected. 

187.  Stresses  in  a  Swing  Span. — The  stresses  for  the  five 
cases  of  loading  will  be  computed  for  the  3oo-ft.  span  (Fig.  197). 
The  assumed  dead  load  is  3,000  Ib.  per  foot  of  span,  or  37,500  Ib. 
per  panel  per  truss.  The  end  panel  load  is  assumed  at  20,000 
Ib.  A  Cooper's  £-40  will  be  taken  for  the  live  load.  The 
impact  allowance  will  be  computed  from  the  formula 

T   -      T         3°° 

1   =  L  j — : — 

*  +  300 


304 


THEORY   OF   FRAMED    STRUCTURES 


CHAP.  VIII 


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SWING  BRIDGES 


305 


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| 

£ 

<N 

5 

M 

00 

ro 

1 

1 

1 

i 

+ 

+ 

j 

M 

^ 

£ 

tS 

^ 

£ 

^ 

^ 

^ 

^ 

b 

^ 

fc> 

20 


THEORY    OF   FRAMED    STRUCTURES 


CHAP.  VIII 


in  which  7  =  impact  stress 
L  =  live  load  stress 
/  =  loaded  length  of  span  causing  the  live-load  stress. 

Since  the  bridge  is  symmetrical  about  the  center,  the  stresses 
will  be  computed  for  the  left  arm  only.  In  Case  III,  this  arm 
is  considered  as  a  span  simply  supported  at  LQ  and  L6.  The 
stresses  are  computed  in  the  ordinary  way.  The  results 
expressed  in  1,000  Ib.  units  are  tabulated  in  Table  I;  the  impact 
stresses  appearing  directly  below  the  corresponding  live-load 
stresses.  Thus  for  the  member  UiUs  the  live-load  stress  is 
—  208,000  Ib.  and  the  impact  stress  is  —139,000  Ib.  For  the 
member  L^U^  the  live-load  and  impact  stress  is  +72,000  Ib. 
when  the  left  segment  of  the  arm  is  loaded;  and  —145,000  Ib. 
when  the  right  segment  of  the  arm  is  loaded. 

In  Cases  IV  and  V  the  reactions,  being  statically  indeter- 
minate, cannot  be  accurately  computed  until  the  truss  has  been 
designed.  In  order  that  a  preliminary  design  may  be  made; 
the  reactions  will  be  tentatively  determined  by  assuming  that 
the  truss  functions  as  a  beam  of  uniform  cross-section,  con- 
tinuous over  three  level  supports.  Only  the  end  reactions  RQ 
are  necessary.  They  are  given  in  Table  II  as  determined  from 
Eqs.  (9)  and  (n)  page  257.  By  this  process  it  is  possible  to 
compute  the  stresses  and  make  a  preliminary  design,  after  which 
the  true  reactions  may  be  determined. 

TALBE  II 


i  Ib.  at 

R0 

1,000  Ib.  at 

Ro 

£1 

+Q-793 

L7 

—  0.064 

L2 

+°-593 

Ls 

—  0.092 

L3 

+0.406 

L, 

-0.094 

L, 

+0.241 

L! 

-0.074 

L, 

+0.103 

Ln 

—  0.040 

188.  Positive  Shear  in  Panel  o-i.  Case  III.— The  influence 
line  abc  for  shear  in  the  panel  is  shown  in  Fig.  197,  from  which 
the  following  criterion  for  maximum  positive  shear  may  be 
developed. 


SEC.  I 


SWING  BRIDGES 


307 


If  PI    =  the  load  in  panel  o-i 

and  P  =  the  total  load  on  the  arm  0-6 

the  criterion  is 


U,      U?      U3 


Uio    Un 


When  the  train  is  moving  to  the  left,  wheel  4  passing  LI  satisfies 
this  criterion  and  the  maximum  shear  in  the  panel  is  +162,000 


308  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VIII 

lb.     The  area  abc  is  62.5  and  the  equivalent  uniform  load  per 
linear  foot  is 

162,000 
q  =  -^-  =  2,500 

Case  IV. — The   influence  line  for  positive  shear  is  defghij 
(Fig.  197),  from  which  the  following  criterion  may  be  developed. 
If  PI  =  the  load  in  panel  o-i 
P%  =  the  load  in  panel  1-2 
P3  =  the  load  in  panel  2-3,  etc. 
the  criterion  is 

793P!  <  2ooP2  +  iSyPs  +  i6sP4  +  i38P5  +  io3P6 
Try  wheel  4  at  L\,  train  moving  to  the  left. 

Wheel  4  approaching  LI 

793Pi  =  793  X  50  =  39>65o        2ooP2  =  200  X  66  =  13,200 

i87P3  =  187  X  56  =  10,472 

i65P4  =  165  X  73  =  12,045 

i38P5  =  138  X  57  =  7,866 

i03P6  =  103  X  50  =  5,150 

48,733 
39,650 <48, 733  therefore  the  shear  is  increasing. 

Wheel  4  having  passed  LI 

793Pi  =  793  X  70  =  55,510        2ooP2  =  200  X  59  =  11,800 

i87P3  =  187  X  43  =  8,041 
i65P4  =  165  X  86  =  14,190 
i38P5  =  138  X  44  =  6,072 
i03P6  =  103  X  50  =  5,150 

45,253 
55,5io>45,253,  therefore  the  shear  is  decreasing. 

When  the  train  is  moving  to  the  left,  wheel  4  at  LI  satisfies  the 
criterion  for  maximum  shear  in  panel  o-i. 

The  shear  may  be  computed  by  scaling  the  length  of  the 
ordinate  in  the  influence  line  for  each  load,  and  taking  the  sum 
of  the  products  of  each  load  and  its  ordinate.  Heretofore  this 
has  been  the  usual  method  of  procedure.  It  requires  that  the 
influence  line  be  drawn  quite  accurately  to  scale,  and  con- 
siderable care  taken  in  scaling  the  ordinates.  The  shear  may 


SEC.  I  SWING   BRIDGES  309 

also  be  determined  by  taking  the  sum  of  each  floor-beam  load 
and  its  corresponding  ordinate  as  follows: 

75.64  X  0.793  =  60.0 
,52. 56  X  0.593  =  31.2 
72.80  X  0.406  =  29.6 
58.20  X  0.241  =  14.0 
48.60  X  0.103  =  5.0 


139.8  =  maximum  shear  in  panel  o-i. 

A  new  and  much  shorter  method,  as  outlined  by  the  author  in 
Engineering  News  Record,  June  9,  1921,  will  now  be  explained. 
If  the  influence  line  defghij  were  straight  from  e  toj,  the  criterion 
for  maximum  shear  in  panel  o-i  would  be  the  same  as  for  Case 
III.  The  difference  between  the  broken  line  efghij  and  a 
straight  line  from  e  toj  is  so  slight  in  this  or  any  similar  truss 
that  the  criterion  for  maximum  shear  in  the  panel  will,  in 
general,  place  the  train  in  the  same  position  or  approximately 
the  same  position,  as  will  the  criterion  for  Case  III.  A  very 
close  approximation  to  the  shear  of  139.8  Ib.  can  be  made  very 
quickly,  by  assuming  the  same  equivalent  uniform  load  in  both 
cases;  or,  in  other  words,  by  assuming  that  the  shears  in  the 
panel  for  the  two  cases  are  directly  proportional  to  the  areas  of 
the  respective  influence  line  diagrams.  These  areas  are 
proportional  to  the  sums  of  their  respective  ordinates,  thus 

area  afcrarea  defghij  1:2. 5:2. 14. 

The  shear  in  the  panel  for  Case  III  was  found  to  be  162  Ib., 
hence  the  shear  for  Case  IV  is 

162.  X^  =  138.3 

2-S 

This  is  a  reasonably  close  approximation  to  the  actual  shear  of 
139.8  Ib.,  previously  determined. 

It  is  now  obvious  that  the  stresses  in  L0Ui  and  L0L2  resulting 
from  positive  shear,  may  be  quickly  found  by  multiplying  the 
stresses  for  Case  III  by  the  ratio  2.14/2.5.  This  ratio  for 
panel  o-i  remains  the  same  for  any  bridge  having  six  equal 
panels  in  each  arm,  irrespective  of  the  length  of  the  panel. 


310  THEORY    OF   FRAMED    STRUCTURES  CHAP.  VIII 

The  stresses  for  Case  III  are  given  in  Table  I  ;  and  the  live  load 
stresses  in  LoUi  and  LoLz  for  Case  IV  are  as  follows: 


L0Ui  =  -2ii  X          =  -180 
=  +135  X^  =  +115 


The  impact  stresses  for  Case  IV  are  determined  in  a  similar 
manner. 

The  influence  line  for  negative  shear  in  panel  o-i  for  Case  IV 
isjklmnop,  and  since  there  is  at  present  no  corresponding  area 
for  Case  III  we  shall  leave  this  question  to  be  considered  later. 

189.  Positive  Moment  about  U3.  Case  III.  —  The  influence 
line  abc  for  moment  about  U%  is  shown  in  Fig.  198. 

Let  PI  =  the  load  on  the  segment  0-3 
and  P  =  the  total  load  on  the  area  0-6 

then  the  criterion  for  maximum  moment  about  Us  is 

p 


Wheel  1  2  at  L3  satisfies  this  criterion,  and  the  maximum  moment 
about  £7  3  is  7,057  ft.-lb.;  hence  the  maximum  live  load  stress 
in  L2L4  is 


Case  IV.  —  The  influence  line  is  defghij,  Fig.  198. 

If  PI  =  the  load  in  panel  o-i 
PI  =  the  load  in  panel  1-2 
PS  =  the  load  in  panel  2-3,  etc. 

the  criterion  for  the  maximum  moment  about  Us,  when  reduced, 
is 


Upon  trial  it  will  be  found  that  wheel  12  at  L3  satisfies  this 


SEC.  I  SWING  BRIDGES  311 

criterion  also.  Multiplying  each  floor-beam  load  by  its  corre- 
sponding ordinate, 

75.20  X  9-475  =  7J3 
51.92  X  19-475  =  ^OI1 
74.52  X  30-45o  =  2,269 
56.56  X  18.075  =  1,022 
48.80  X  7-725  =  _377 

5,392  =  maximum  moment  about  U%. 

The  maximum  tensile  stress  in  L2L4  is 

=  +179-7 
30 

This  value  may  be  closely  approximated  from  Case  III  as 
follows:  The  ratio  of  the  sum  of  the  ordinates  in  Case  IV,  to  the 
sum  of  the  ordinates  in  Case  III,  is 

^  -  0.756 
112.5 

and  235  X  0.756  =  +178  =  stress  in  I^L4. 

190.  Negative  Shear  in  Panel  o-i.     Case  IV.  —  We  are  now 

prepared  to  consider  the  negative  shear  in  panel  o-i,  when  the 
right  arm  6-12  is  loaded.  The  influence  line  isjklmnop  (Fig. 
197),  and  the  criterion  developed  therefrom  is 

+  347*11  +  40^12. 


There  are  two  positions  of  the  train,  when  moving  to  the  left, 
which  satisfy  this  criterion  ;  namely,  wheel  1  1  at  L%  and  wheel  8 
at  Lg.  The  maximum  shear,  occurring  when  the  train  is  in  the 
latter  position,  is  —22.8  lb.;  as  found  by  taking  the  sum  of  the 
products  of  each  floor  beam  load  and  its  corresponding  ordinate. 
This  value  may  be  closely  approximated  by  the  proportionate 
method,  as  follows:  Since  the  influence  Imejklmnop  (Fig.  197) 
has  its  longest  ordinate  at  the  center  of  the  arm,  it  will  be 
compared  with  the  influence  line  abc  (Fig.  198),  which  also  has 
its  longest  ordinate  at  the  center  of  the  arm.  The  latter 
influence  line  is  for  the  moment  at  Us  for  Case  III,  and  the 
moment  is  7,057.  The  ratio  of  the  sum  of  the  ordinates 


3I2 


THEORY    OF   FRAMED    STRUCTURES  CHAP.  VIII 


jklmnop  (Fig.  197)  to  the  sum  of  the  ordinate  abc  (Fig.  198)  is 

—0.364 

?  =  —0.00324 
112.5 

and          7,057  X  —0.00324  =  —22.9  =  shear  in  panel  o-i. 
i     U_2_U3     tk     U5    U&     U?     Us 


F 1 0.200 


This  quantity  represents  also  the  maximum  negative  reaction 
at  LO  when  the  right  arm  is  loaded,  from  which  all  the  stresses 


SEC.  I  SWING  BRIDGES  313 

may  be  determined  as  given  in  the  designated  column  of  Table  I. 
191.  Shear  in  Panel  1-2.  —  The  influence  lines  are  shown  hi 
Fig.  199.  The  stresses  in  UiLz  for  Case  III,  as  given  in  Table 
I,  are  —  n  and  +  140;  from  which  the  stresses  for  Case  IV  may 
be  found  as  follows: 


0-593  +  0.406  +  0.241  +  0.103  =  + 

0.667  +  0.500  +  0.333  +  °-l67 

The  stresses  for  Case  IV,  determined  by  the  exact  method,  are 
—  15  and  +113  respectively. 

192.  Shear  in  Panel  2-3.  —  The  influence  lines  are  shown  in 
Fig.  200.     The  stresses  in  L2C/3  for  Case  IV,  determined  by  the 
proportionate  method  are 

x  0,207+0407  .  +48 
0.167  +  0.333 

_82  x  +  0.241+0.103  =  _6l    . 

0.500  +  0.333  +  0.167 

The  stresses  for  Case  IV,  found  by  the  exact  method,  are  +48 
and  —60  respectively. 

193.  Shear  in  Panel  3-4.  —  The  influence  lines  are  shown  in 
Fig.  201.     The  stresses  in  UsL^  for  Case  IV,  determined  by  the 
proportionate  method,  are 


-82  X  °'2°7       °'4°7       °'594  =  -99 
0.167  +  0.333  +  0.500 

+39  X  9^41+0103       + 
0.333  +  0.167 

The  stresses  for  Case  IV,  determined  by  the  exact  method  are 
—  101  and  +26  respectively. 

194.  Shear  in  Panel  4-5.  —  The  influence  lines  are  shown  in 
Fig.  202.  The  stresses  in  L±U5  for  Case  IV,  determined  by  the 
proportionate  method,  are 


X  °'2°7      °'4°7      °'594      °-7g  =  +165 

0.167    +  0.333    +  0.500   + 


THEORY  OF   FRAMED   STRUCTURES 


CHAP.  VIII 


The  stresses  for  Case  IV,  determined  by  the  exact  method,  are 
+  167  and  —5  respectively. 

195.  Shear  in  Panel  5-6. — The  influence  lines  are  shown  in 

U3      U4      U5      U6     U-T      U8     U9 


Fig.  203.     The  stress  in  U5L6  for  Case  IV,  determined  by  the 
proportionate  method,  is 

-2ii  X  °-2°7  +  0407  +  Q.594  +  0-759  +  0-897  =  _2 
0.167  +  0-333  +  0.500  +  0.667  +  0.833  " 


SEC.  I 


SWING  BRIDGES 


315 


The  stress  for  Case  IV,  determined  by  the  exact  method,  is 
-241. 
196.  Moment  about  1/2. — The  influence  lines  are  shown  in 

Us    Us     UIQ    Uu 


'W       FIG.204 

Fig.  204.     The  stress  in  UiU*  for  Case  IV,  determined  by  the 
proportionate  method,  is 

,  81.8 

—  208  X  -     -  =   —170 
100 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VIII 


The  stress  for  Case  IV,  determined  by  the  exact  method,  is 
-172. 
197.  Moment  about  L4. — The  influence  lines  are  shown  in 


U,      U? 


U4     U5     U& 


U8     U9     Uie    U,, 


Lo 


CaseHI 


CaselY 


iaseUI 


FIG.  2 06 


LII        t-12 


Fig.  205.     The  stress  in  USU5  for  Case  IV,  determined  by  the 
proportionate  method  is 


-208 


100 


=    -I32 


SEC.  I 


SWING  BRIDGES 


317 


The  stress  for  Case  IV,  determined  by  the  exact  method,  is 

-134- 

198.  Moment  about  U5. — The  influence  lines  are  shown  in 
Fig.  206.  Whenever  each  arm  has  six  or  more  equal  panels,  the 
influence  line  for  Case  IV  shows  a  reversal  of  stress  in  one  or 
more  chord  members  of  the  loaded  arm  adjacent  to  the  center 
support.  This  phenomenon  is  explained  in  connection  with 


Fig.  207,  where  a  is  the  distance  from  R\  to  the  point  of  zero 
bending  moment;  then 

aRi=   P(a  -  kl) 

fromEq.  9,  page  257 

p 

4 
hence  a  =  -    !— - 

For  any  position  of  P,  the  limits  of  k  are  o  and  i ;  hence  the 
limits  of  a  are  %l  and  /.  It  is  clear,  therefore,  that  if  any  panel 
point  experiences  a  negative  moment  from  the  influence  load; 
the  distance  of  that  panel  point  from  the  center  support  must 
be  less  than  J£  of  the  arm  length.  If  the  panels  are  of  equal 
length  this  condition  can  occur  only  when  there  are  six  or  more 
panels  in  each  arm. 


3*8  THEORY   OF   FRAMED    STRUCTURES  CHAP.  VIII 

The  stresses  for  Case  IV  cannot  be  determined  by  the  pro- 
portionate method  as  heretofore,  on  account  of  the  dissimi- 
larity of  the  influence  line  diagrams.  In  such  instances  the 
equivalent  uniform  load  may  be  approximated  by  the  use  of 
equivalent  uniform  load  table,  page  206.  For  the  right 
segment,  l\  =  25,  /2  =  61.5  and  the  equivalent  uniform  load  is 
2.7;  the  area  is  464  and  the  stress  in  L^L6  is 


= 


30 

For  the  left  segment,  it  will  be  sufficient  to  call  l\  —  li  =  30,  and 
the  equivalent  uniform  load  is  2.88,  hence  the  stress  is 
-38  X  2.88  _ 
~£~ 

199.  Moment  about  L6.  —  The  influence  line  for  Case  IV  is 
shown  in  Fig.  208.     There  is  no  corresponding  influence  line  for 
Case  III.     Since  the  influence  is  symmetrical  about  the  center, 
the  stress  in  U^U^,  when  the  left  arm  is  loaded,  is  the  same  as 
previously  determined  when  the  right  arm  was  loaded. 

200.  Case  V:Both  Arms  Loaded.     Broken  Loads.  —  By  re- 
ferring to  the  influence  lines  for  the  continuous  condition  in  Figs. 
197,  198,  204  and  205,  it  is  apparent  that  the  stresses  in  the  mem- 
bers there  considered  will  be  less  for  Case  V  than  for  Case  IV; 
since  any  load,  brought  on  to  the  right  arm  while  the  left  arm  is 
loaded,  will  decrease  the  stress  because  the  influence  areas  on 
opposite  sides  of  the  center  have  opposite  signs.     In  Figs.  199, 
200,  201,  202,  203,  206  and  208  the  conditions  are  different;  since 
loads  on  the  right  arm  and  the  left  segments  of  the  left  arm  con- 
spire, and  the  live-load  stress  in  any  instance  is  the  sum  of  the 
live  stresses  as  given  for  Case  IV.     Consider  the  member  U&Lt, 
for  example,  illustrated  in  Fig.  201.     The  live-load  stress  is  —  99 
when  the  left  segment  of  the  left  arm  is  loaded,  and  —  30  when 
the  right  arm  is  loaded;  hence  if  a  train  approaches  on  each  arm, 
the  maximum  live  load  stress  is  —129,  as  given  in  the  column 
for  broken  loads.     No  impact  is  added  when  broken  loads  are 
considered.     Specifications  are  not  usually  clear  on  the  ques- 
tion of  broken  loads.     If  the  location  of  the  bridge  is  near  a 
large    freight    terminal,    it  is   conceivable  that  trains  might 


SEC.  I  SWING  BRIDGES  319 

occasionally  approach  simultaneously  from  both  ends  of  the 
bridge. 

Continuous  Loads.  —  If  Figs.  199,  200,  201,  202,  are  taken  con- 
secutively, it  is  apparent  that  the  positive  influence  line  area  for 
Case  IV  is  decreasing,  and  in  Fig.  201  it  becomes  less  than  the 
negative  area  for  the  right  arm;  the  difference  in  areas  being 
2.65.  Therefore  the  stress  in  UsL*  for  Case  V  will  be  greater 
than  for  Case  IV. 

The  stress  may  be  approximated  as  follows:  Consider  that 
the  train  moving  to  the  left  covers  the  whole  span.  Assume 
that  the  engine  covers  the  segment  ab  and  that  the  stress  in 
Z73Z,4,  on  account  of  the  engine,  is  the  same  as  in  Case  IV  or 
—  99  Ib.  The  shear  in  panel  3-4,  on  account  of  uniform  train 
load  of  2,000  Ib.  per  linear  foot.,  from  b  toe  is  —  2.65  X  2  =  —5.3 
Ib.,  and  the  stress  in  U^L^  is  —5.3  X  1.3  =  —6.9.  Hence  the 
stress  for  Case  V  is  —  99  —  6.9  =  —  105.9.  Similarly  the  stress 
in  L4£/5  is  +165  +  (7.6  X  2  X  1.3)  =  +184.8. 

20  1.  Negative  Shear  in  Panel  5-6.  —  There  is  no  positive 
area  in  the  influence  line  for  the  continuous  condition  of  Fig.  203. 
The  train  must  be  moving  to  the  left,  if  the  engines  are  to  be  on 
the  segment  ab  ;  followed  by  a  uniform  train  load  on  the  segment 
be.  Since  the  segment  ab  is  considerably  longer  than  the  length 
of  the  two  engines  ;  the  shear  in  panel  5-6  for  Cases  III  and  IV 
would  be  considerably  less,  if  the  engines  were  moving  toward 
the  left  instead  of  in  the  opposite  direction.  We  shall  take 
this  difference  into  consideration,  by  computing  the  negative 
shear  in  panel  5-6  when  the  train  is  moving  to  the  left.  This 
occurs  when  wheel  16  is  at  L^  and  the  shear  (Case  III)  is  —  151.7 
Ib.;  which  is  considerably  less  than  —162  Ib.,  when  the  train 
is  moving  to  the  right.  Taking  the  same  ratio  of  ordinates  as 
before,  the  shear  for  Case  V  is 


—  9.1   X  2  =    —     l8.2 

-192. 

The  stress  in  U&LG  is  —192.0  X  1.3  =  —249.6. 

202.  Moment  about  L6.—  The  stress  in  U5U6  for  Case  IV 
was  found  to  be  +115,  one  area  being  loaded.     When  the  other 


320  THEORY    OF   FRAMED    STRUCTURES  CHAP.  VIII 

arm  is  covered  with  a  uniform  train  load  of  2,000  Ib.  per  foot,  the 
additional  stress  is  +91,  hence  the  stress  for  U5U&  for  Case  V  is 
+  206. 

203.  Dead  Load :  Bridge  Open.     Case  I. — The  panel  loads 
at  each  end  are  20,000  Ib.  and  all  others  37,500  Ib.     Each  truss  is 
balanced  on  the  center  support  and  the  stresses  are  statically 
determinate.     They  are  given  in  Table  I. 

204.  Dead  Load :  Ends  Raised.     Case  II. — The  maximum 
negative  reaction  on  account  of  live  load  was  found  to  be  —  22.9, 
to  which  must  be  added  — 15.3  per  impact,  or  a  total  of  —38.2. 
Hence,  if  the  positive  uplift  at  each  end  is  38.2  or  greater,  there 
will   be   no   hammering   of  one  end  when  the   train  covers 
the  opposite  arm  of  the  bridge.     It  will  be  assumed  that  the 
machinery  parts  are  to  be  designed  and  adjusted  so  that  the 
end  wedges,  when  driven,  will  exert  an  upward  pressure  of  50. 
Since  there  is  a  dead  load  of  20  at  the  end  panel  point,   the 
resultant  or  net  positive  end  reaction  is  30;  and  the  resulting 
stresses  are  given  in  Table  I.     If  the  truss  were  treated  as 
fully  continuous,  the  end  reaction  would  be  86.4  instead  of  50 ^ 
and  a  heavier  and  more  expensive  lifting  device  would  be 
required. 

205.  Combinations. — As  previously  explained,  Case  I  is  com- 
bined with  either  Case  III  or  Case  V ;  and  Case  II  is  combined 
with  either  Case  IV  or  Case  V.     Only  two- thirds  of  the  dead- 
load  stress  is  taken,  when  dead-load  and  live-load  stresses  have 
opposite  signs.     Many  specifications  are  not  clear  upon  the 
question  of  stress  reversals.     In  treating  reversals,  each  com- 
bination  should   be   considered   separately;   i.e.,   the   largest 
positive  stress  of  one  combination  should  not  be  considered  with 
the    largest    negative    stress    of    another    combination.     The 
members  have  been  proportioned,  and  the  gross  cross-sectional 
areas  given  in  Table  III. 

PI 

206.  Reactions  from  Williot  Diagram. — The  quantities  -j 

in  Table  III,  when  divided  by  the  modulus  of  elasticity  E,  are 
the  strains  in  the  various  members  when  the  center  reaction 
is  removed  and  the  truss,  supported  at  o  and  12,  carries  a  load 
of  i  Ib.  at  joint  6  (Fig.  209.)  The  Williot  diagram  is  drawn 


SEC.  I 


SWING  BRIDGES 


32I 


PI 

in  Fig.   210  by  using  the  quantities  ~^  to  represent  strains. 
The  quantities  d  which  are  proportional  to  the  deflections  are 

U,       2        1        4-        5        6        7        8        9        '~      " 


470  455  414  342 

r/# 


?38     /?/ 


/?/    ?3&   We    414    455 

FIG.  209. 

indicated  in  Fig.  209.  The  deflection  at  the  center  has  been 
checked  in  Table  III,  where  P  and  UQ  necessarily  have  the  same 
numerical  values;  but  it  should  be  remembered  that  P  is 
measured  in  pounds,  while  u&  is  a  ratio.  LO 

It  is  clear  from  Maxwell's  Theorem 
that  if  i  Ib.  at  joint  6  causes  the  deflec- 
tion di  =  121  at  joint  i;  then  i  Ib.  at 
joint  i  will  cause  the  deflection  d\  =  121 
at  joint  6.  Hence,  with  i  Ib.  at  joint 
i,  the  reaction  at  joint  6,  when  the  truss 
is  continuous  over  the  three  supports,  is 

R6  =  —  =  0.258 
470 

and  from  statics 
R   =  i.o  X  275  -  (0.258  X  150) 
300 


=  0.788 


The  reactions  at  the  left  end,  determined 
in  the  same  manner,  are  given  in  Table 
IV. 

If  the  accurate  reactions  in  Table  IV 
are  compared  with  the  assumed  reactions 
of  Table  II,  it  will  be  apparent  that  the 
differences  are  comparatively  small.  The 
greatest  percentage  of  error  occurs  when 
the  load  of  i  Ib.  is  at  the  center  of  either 
arm.  This  comparison  gives  a  fair  idea 
of  what  error  may  be  expected,  when  the 
truss  is  assumed  to  be  a  beam  of  constant  moment  of  inertia; 
and  no  recognition  is  made  of  deflection  due  to  shear. 


322 


THEORY   OF   FRAMED    STRUCTURES 
TABLE  III 


CHAP.  VIII 


Member 

Length, 
inches 

Area, 
square  inches 
A 

Stress   in 
pounds 
P 

PI 

A 

w6 

Pud 
A 

L*-Vi 

469 

32.5 

—  0.652 

-9.40 

—  0.652 

+6.13 

Ui-  U2 

300 

30.5 

-0.833 

—  8.20 

-0.833 

+6.84 

U2-  Us 

300 

30.5 

-0-833 

—  8.20 

-0.833 

+6.84 

Us-  Ui 

300 

30.5 

-1.666 

—  16.40 

-1.666 

+  27-35 

Ui~  f/5 

300 

30.5 

—  1.666 

—  16.40 

-  1  .  666 

+  27-35 

Z/6-  Us 

300 

68.2 

—  2.500 

—  10.99 

—  2.500 

+  27.46 

it-ii 

300 

19.8 

+0.416 

+6.32 

+0.416 

+  2.63 

L,-LZ 

300 

19.8 

+0.416 

+6.32 

+0.416 

+  2.63 

LZ-LS 

300 

26.5 

+  1.250 

+  14.14 

+  1.250 

+  17.68 

Ls  -Li 

300 

26.5 

+  1.250 

+14.14 

+  1.250 

+  17.68 

Li-L$ 

300 

44-5 

+  2.083 

+14-05 

+  2.083 

+  29-25 

LS-LG 

300 

44-5 

+  2.083 

+14-05 

+  2.083 

+  29-25 

Ui-  Lz 

469 

19.8 

+0.652 

+15-44 

+0.652 

+  10.08 

Lz-  Us 

469 

19.8 

—  0.652 

—  15-44 

—0.652 

+  10.08 

Uz-  Li 

469 

32.5 

+0.652 

+9.40 

+0.652 

+6.13 

Li-  Uf, 

469 

44-5 

-0.652 

-6.88 

-0.652 

+4-49 

Us-Ls 

469 

68.2 

+0.652 

+4.48 

+0.652 

+  2.92 

Ui-Lj 

0 

Uz-  L2 

0 

U3-  L3 

0 

Ui-Li 

0 

u*,-Ls 

o 

U.-Lt 

0 

dG  =  Z 


PuJ, 


234-79 

2 


=  469.58 


TABLE  IV 


lib. 

at 

Ro 

i  Ib. 
at 

Ro 

Li 

+0.788 

Li 

—0.068 

Lt 

+0.581 

Ls 

—  o.  108 

L3 

+0.386 

L, 

—  o  .  114 

L* 

+0.226 

LIQ 

—0.088 

L, 

+0.099 

Ln 

—  0.046 

SEC.  I  SWING  BRIDGES  323 

The  reactions  for  the  preliminary  design  might  be  obtained 
by  assuming  that  all  members  have  the  same  cross-sectional 
area,  which  may  be  taken  as  i  sq.  in.;  and  constructing  a 
Williot  diagram. 

The  combinations  of  stresses  for  Cases  I  and  III  determine 
the  sizes  of  nearly  all  members,  except  the  end  post  and  chord 
members  adjacent  to  the  center  support.  Since  the  stresses 
for  these  cases  are  statically  determinate,  they  might  be  used 
in  making  an  estimate  of  the  sizes  of  the  members;  and  their 
areas  used  in  constructing  a  Williot  diagram. 

The  continuous  girder  formulas  give  such  satisfactory 
results  that  a  re-design  is  seldom  necessary,  except  in  a  long 
span  and  then  only  for  a  few  members  adjacent  to  the  center 
support. 

SEC.  II.    RIM-BEARING  SWING  BRIDGES 

207.  The  trusses  in  a  rim-bearing  swing  bridge  are  supported 
by  a  large  circular  girder,  which  rotates  with  the  span.  The 
girder  rests  on  conical  rollers,  usually  about  18  in.  in  diameter; 
and  as  many  rollers  are  used  as  the  circumferential  length  of 
the  girder  will  permit,  in  order  to  give  as  many  bearings  for 
the  girder  as  possible,  and  thereby  minimize  the  deflection. 
The  trusses  may  rest  directly  upon  the  circular  girder  or  drum, 
as  in  Fig.  2116.  This  arrangement  is  undesirable,  because  it 
does  not  give  an  equal  distribution  of  the  load  to  the  rollers. 
A  better  arrangement  is  shown  in  Fig.  2110,  where  the  truss 
loads  are  distributed  to  the  drum  at  8  points  instead  of  at  4 
points,  as  in  the  previous  case.  This  arrangement  gives  a 
more  even  bearing.  In  the  diagrams  here  shown,  the  center 
pivot  receives  neither  dead  nor  live  load.  It  is  better  to  frame 
the  structure  so  that  from  15  to  20  per  cent  of  the  load  is 
transmitted  to  the  center  pivot  through  radial  girders.  In  any 
case  each  truss  is  supported  at  two  points  over  the  circular 
pier,  as  illustrated  in  Fig.  2im. 

The  reactions  for  the  center-bearing  bridge  of  the  previous 
section  were  determined  by  assuming  that  the  span  functions 
as  a  beam  of  constant  cross  section,  continuous  over  three  sup- 
ports; no  allowance  being  made  for  deflection  due  to  shear. 


3  24 


THEORY    OF    FRAMED    STRUCTURES 


CHAP.  VIII 


It  was  shown  that  the  reactions  thus  computed  by  continuous 
girder  formulas,  compared  very  favorably  with  the  true  reac- 
tions determined  after  the  design  had  been  made.  Such  is  not 
the  case  when  the  continuous  girder  formulas  of  Art.  167  are 


CossET 


^      TV 


CO 


PIG.  211. 


applied  to  a  swing  span  on  four  supports.  These  formulas 
give  a  negative  reaction  at  LI  and  a  positive  reaction  at  L\^ 
when  the  left  arm  is  loaded.  This  live  load  negative  reaction, 
when  the  impact  factor  is  added,  is  in  many  instances,  numer- 
ically greater  than  the  dead  load  positive  reaction;  indicating 
that  a  live  load  over  the  left  arm  would  lift  the  truss  from  its 


SEC.  I  SWING  BRIDGES  325 

support  at  LI.  This  assumption  is  not  justified  either  by 
exact  analysis,  made  after  a  truss  is  designed;  or  by  observed 
data  taken  after  erection. 

208.  Partial  Continuity.  Equal  Moments  at  Center  Sup- 
ports. —  It  will  be  assumed  that  the  diagonal  bracing  in  panel  6-7 
is  so  light  that  no  appreciable  shear  can  be  transmitted  through 
this  panel.  The  shear  in  the  panel  is  assumed  zero  under  any 
condition  of  loading;  hence  RT  =  —R  13  for  loads  on  the  arm  0-6, 
and  R&  =  —Ro  for  any  loads  on  the  arm  6-13.  The  formulas 
given  in  Art.  172  are  applicable  where  a  =  20/150;  hence  for  an 
influence  load  of  one  pound 


and 

k  —  k3 

T>  K  K 

13      "      7~»~ 
4-0 

The  five  cases  of  loading  to  be  considered  are  the  same  as  for 
a  center-bearing  bridge.  The  influence  lines  for  shear  in  panel 
o-i  for  Cases  III  and  IV  are  shown  in  Figs.  21  id  and  2iie,  re- 
spectively. If  the  live  load  is  an  £40,  the  stresses  for  Case  III 
will  be  the  same  as  for  the  center-bearing  bridge  given  in  Table 
I.  The  stresses  for  Cases  IV  and  V  may  be  found  by  the  pro- 
portionate method.  For  example,  the  stress  in  LoUi  for  Case 
III  is  —2ii,  hence  the  stress  for  Case  IV  is 


-211  X  •  -  -  .  .         _         g 


The  influence  lines  are  drawn  and  the  stresses  computed  for  all 
the  members,  in  precisely  the  same  manner  as  for  the  centre- 
bearing  bridge.  It  may  be  noted  that  in  this  particular  problem 
the  positive  shears  are  greater,  and  the  negative  shears  less,  in 
the  rim-bearing  type  than  in  the  center-bearing  type.  The 
stresses  in  some  members  will  be  greater,  and  in  others  less 
than  in  the  center-bearing  bridge.  If,  however,  an  independent 
design  is  made  for  the  rim-bearing  type,  a  comparison  will  show 
no  appreciable  difference  in  the  two  designs.  For  this  reason 
it  is  a  common  practice  to  disregard  the  center  panel  when  the 
stresses  are  computed.  The  diagonals  in  the  center  panel  are 


326  THEORY   OF   FRAMED    STRUCTURES  CHAP.   VIII 

light  adjustable  members,  which  serve  only  to  provide  stability 
to  the  structure  when  open,  and  resisting  a  longitudinal  wind 
pressure. 

After  the  trusses  have  been  designed,  a  sufficiently  exact 
analysis  of  the  reactions  may  be  made  by  omitting  the  bracing 
in  the  center  panel;  removing  the  center  supports,  and  drawing 
a  Williot  diagram  for  one  pound  loads  placed  at  L6  and  £7. 


INDEX 


Arch,  three  hinged,  75,  78,  109 
Area,  moment  method  for  deflections, 
208    ' 

B 

Bay,  109 
Beam  defined,  83 
Beams  under  moving  loads: 

criterion  for  maximum  bending 

moment,  134,  143 
shear,  146 
influence  line  for  bending  moment, 

132 

reaction,  130 
shear,  132 

illustrative  problem,  138 
point  of  greatest  maximum  bend- 
ing moment,  143 
Beams  under  stationary  loads: 

bending  moment  determined 
from  location-direction  dia- 
gram, 84 

shear  and  bending   moment  dia- 
grams, 86-99 
concentrated   loads,  87 
concentrated    and    uniform 

loads,  96 
uniform  loads,  90 
continuous,  240,  248-256 

Clapeyron's    theorem  of   three 

moments,  268 
coefficients    for    pier  reactions; 

267 

four  equal  spans,  266 
four  supports,  263 
general  expressions,  257 
three  equal  spans,  264 
two  unequal  spans,  259 
continuous  in  foundations,  273 
projection   not  limited  by  site, 

275 

projection  at  one  end  limited  by 
site,  277 


Beams  under  stationary  loads: 

continuous  projection  at  both  ends 

limited  by  site,  279 
Beams : 

deflection  of,  208 
cantilever,  232 
maximum  deflection,  215 
point  of  maximum  deflection, 

218 

reciprocal  deflections,  231 
several  concentrated  loads,  226 
simple    beams,    uniform   cross 

section,  212 
uniform  load,  221 
uniform  and  concentrated  load, 

229 
uniformly    varying    load,    224, 

225 
varying  cross  section,  235 

depth,  236 

with  cover  plates,  237 
partially  continuous,  270 
no  shear  transmitted,  270 
no  moment  transmitted,  271 
restrained,  240 

at    both     ends,      concentrated 

load,  243 
uniform  load,  247 
at  one  end,  concentrated  load, 

241 

uniform  load,  242 
Bending  moment: 
defined,  83 

determined    from   location-direc- 
tion diagram,  84 
diagrams,  86-89 
Bent,  109 
Bridge  trusses: 

Baltimore,  129,  184 

criterion   for   chord    members, 

4& 

for  web  members,  190,  194 
influence  lines  for,  191 
Parker,  129,  169 


327 


328 


INDEX 


Bridge    trusses: 

Parker,  criterion  for  chord  mem- 
ber, 149 

for  web  member,  172,  177 
influence  line  for  web  member, 

172,  177 

stress  in  web  member,  169 
tension     in      vertical      when 

counters  are  used,  182 
Pennsylvania,  130,  197 

influence  lines  for,  198 
Pratt,  129,  146 

chord  stresses  with  odd  number 

of  panels,  167 
criterion    for    chord    member, 

149 

for  web  member,  152,  157 
influence   line  for  chord  mem- 
ber, 146 

for  shear  in  panel,  151 
for  web  member,  151 
stresses  in  a  200  ft.  truss,  160 
moving  loads,  130 
Standard  types,  128 
Warren,  129 


Camber,  298 

Cantilever  bridge,  80 

Clapeyron's  theorem  of  three  moments, 

268 
Cooper's  standard  train  loads,  137 

moment  table  for,  136 
Counters^  166 
Couple,  5 
Culmann's  method  of  sections,  69 


Deflection  of  beams,  See  Beams, 
of  trusses,  See  Trusses. 


Equilibrium : 

equations  of,  12 

polygon,    See    Location-direction 
diagram. 


Equivalent  uniform  loads,  200 
table  for,  206 


Force : 

component  of,  6 

denned,  4 

direction  of,  5 

elements  of,  5 

location  of,  5 

magnitude  of,  5 

moment  of,  1 1 

polygon,  See  Magnitude-direction 

diagram. 

rectangular  components  of,  10 
sense  of,  5 
Forces : 

concurrent,  5 
composition  of,  6 
coplanar,  5 
concurrent,  14 

algebraic  method,  14 

combinations  of  elements,  22 

general  case,  17 

graphic  method,  17 

illustrative  problem,  14, 19,  20 

number  of  independent  equa- 
tions possible,  19 

problems,  23 

special  case,  21 
non-concurrent,  non-parallel,  30 

algebraic  method,  38,  48,  50, 

53 

combinations  of  elements,  38 
general  case,  32 
graphic  method,  41,  48,  52,  54 
groups  of  equations,  33 
illustrative  problem,  31,  38- 

54 
number  of  independent 

equations  possible,  33 
one  unknown  location,  37 
problems,  54 
transformed  systems,  30 
parallel,  24 

algebraic  method,  27 
combinations  of  elements,  30 
general  considerations,  24 
general  case,  26 


INDEX 


329 


Forces : 

coplanar  parallel  graphic  method, 

47 

illustrative  problem,  27 
number   of   independent 
equations  possible,  30 

external,  62 

internal,  62 

non-concurrent,  5 

parallelogram  of,  6 

resolution  of,  6 

resultant  of,  6 

system  of,  5 

triangle  of,  6,  8,  9 


Impact,  159 
Inclined  loads,  81 

Influence     lines,     See     Structure     in 
question. 


Joints,  methods  of,  63 
L 

Location-direction  diagram,  9 

drawn  through  given  points,  45 

M 

Magnitude-direction  diagram,  9 
Maxwell's      theorem      of      reciprocal 

deflections,  231,  299 
Mechanics,  i 
Mental  attitude,  2 
Methods,  2 

'Mill  building,  column  reactions,  118 
Moving  loads,  130 


0 


Odd  number  of  panels,  167 


Parabola,  91 

methods  of  constructing,  93,  94 
Portal  frames,  102-107 
Purlins,  109 


Rankine's  method  of  sections,  68,  69 
Rigid  body  defined,  60 
Ritter's  method  of  moments,  69 
Roof  trusses: 

bracing,  no,  in 

conclusions,  126 

covering,  in 

dead  loads,  in 

design  stresses,  125 

Fink,  73,  108 

Howe,  108 

monitor,  109 

normal  wind  pressures,  114 

purlins,  109 

reactions  of,  82,  116 

saw  tooth,  1 08 

snow  load,  112,  125 

spacing  of,  in 

standard  types,  108 

supported  on  columns,  116 
on  walls,  115 

Warren,  108 

weights  of,  112 

wind  analysis,  115,  120 
load,  113,  125 

stress   diagram   unnecessary, 
124 

S 

Sections,  method  of,  68 
Shear,  denned,  83 

diagrams,  86-99 
Stevin,  6 
Stress  diagram,  65 

and  strain,  282 

defined,  62 

determined  by  method  of  joints, 

63 

of  sections,  68 
for  stationary  loads,  67 
Structural  engineer's  parabola,  91 
Swing  bridges,  301 

center  bearing,  301 
combinations  of  stresses,  303 
conditions  of  loading,  301 
reactions  from  formula,  306 
from  Williot  diagram,  320 


330 


INDEX 


Swing  bridges,  short  method  for  com- 
puting stresses  in,  309 
stresses  in,  303 
rim  bearing,  323 


Trusses : 

assumptions  in   connections 
with,  6 1 

camber  of,  298 

denned,  61 

deflection    of: 

algebraic  method,  283 
due  to  temperature,  297 


Trusses: 

deflection  of: 

graphic  method  (Williot  dia- 
gram), 288 
first  solution,  289 
second  solution,  293 
third  solution,  295 
Maxwell's  theorem,  299 
stability  of,  61 


Varignon's  theorem,  u 

W 
Williot  diagrams,  288,  290,  293,  296 


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